## Your number is, in fact, not bigger!

**Moderators:** jestingrabbit, Moderators General, Prelates

### Re: Your number is, in fact, not bigger!

Just on the busy beaver function, we really have no idea how fast it grows, but it basically works like the winners of the original thread, it just gets the proofs of termination for free. A C program can be converted into a Turing machine with a reasonable increase in storage space. All of our notations could be implemented in C in a few thousand characters, so it's not unreasonable to say that they could be encoded as Turing machines with a few million states. BB(9) greatly exceeds a few million, so BB(BB(9)) excedes any value you could conceivably write using computable functions. This is all very vague, but it's safe to say we'll never get anywhere near BB(BB(9)) using computable functions.

Looking forward to exploring WarDafts notation later tonight.

Looking forward to exploring WarDafts notation later tonight.

addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.

- Vytron
**Posts:**429**Joined:**Mon Oct 19, 2009 10:11 am UTC**Location:**The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P

### Re: Your number is, in fact, not bigger!

WarDaft wrote:But then:

n,0,{0,0} = ω^ω*ω = ω^(ω+1)

n,0,0,{0,0} = ω^(ω+1)*ω = ω^(ω+2)

n,{0},{0,0} = ω^ω*ω^ω = ω^(ω*2)

n,0,{0},{0,0} = ω^(ω*2)*ω = ω^(ω*2+1)

n,0,0,{0},{0,0} = ω^(ω*2+1)*ω = ω^(ω*2+2)

n,{0,0,0} = n,<0,>{0},{0,0} = ω^(ω*3), not ω^ω^ω

Remember the exponent laws, ω^a*ω^b = ω^(a+b)

It's meant to reduce the smaller ordinal first, so we have:

n,0,{0},{0,0}

0,{0} is the smaller string, so it reduces to:

n,{0},{0}...{0},{0},{0,0}

And then you make X copies of that:

n,{0},{0}...{0},{0},{0,0},{0},{0}...{0},{0},{0,0},{0},{0}...{0},{0},{0,0} (3 copies are shown but they should be n)

Not n,0,{0},{0,0} = n,{0},{0,0},{0},{0,0}...{0},{0,0}

WarDaft wrote:Of course, the [-] notation totally compensates for this, because it doesn't even matter if you start with the function (+1) or f(a) = ζa, the [-] notation on its own is strong enough to reach φ(1@5).

Wait, what? This is all about nested concatenation? I don't know if it's depressing or hilarious how I've basically wasted all my time writing, rewriting and polishing my base notation, which doesn't seem to matter at this point

Anyway, you know my notation runs out of fuel before ψ(Ω^Ω^Ω^Ω), I know yours can be extended to go beyond that, so I'm going to end losing the +1ing war.

So what I'm going to do is switching gears and implement Illusory Space in a new, nice way, which you will not even notice is there!

Everything up to this point in my last post applies:

n,1 = n,[0-0-0-0] for n -s

n,1 = φ(1@ω) so that k,1 = φ(1@k)

And that's the Step 0 Of Level 0 of... Wait, I don't think I have to name things like that, or use colored brackets, I just have to define:

(n),x = n,x

If x doesn't have an integer higher than 0.

Otherwise:

(n,0),x enumerates the fixed points of a -> (a),x

(n,0,0),x enumerates the fixed points of a -> (a,0),x

(n,0,0...k 0s...),x enumerates the fixed points of a -> (a,0,0,...k-1 0s...,0,0),x

(n,{0}),x enumerates the fixed points of a -> (a,0,0,0),x for n 0s

(n,0,{0}),x enumerates the fixed points of a -> (a,{0},{0},{0}),x for n {0}s

And so I send:

(3,{0}),1

For φ(1@(ω2))

mike-l wrote:Just on the busy beaver function, we really have no idea how fast it grows, but it basically works like the winners of the original thread, it just gets the proofs of termination for free. A C program can be converted into a Turing machine with a reasonable increase in storage space. All of our notations could be implemented in C in a few thousand characters, so it's not unreasonable to say that they could be encoded as Turing machines with a few million states. BB(9) greatly exceeds a few million, so BB(BB(9)) excedes any value you could conceivably write using computable functions. This is all very vague, but it's safe to say we'll never get anywhere near BB(BB(9)) using computable functions.

Yeah, apparently for Busy Beavers functions what values you reach doesn't really matter, it's all about finding their lowerbounds. Here's this page with lowerbouds, and we have this thing:

BB(3750) > f{ψ_Ω_1(ε_Ω_{ω+1})}(683) (in the FGH)

Which is a number that completely obliterates everything I've ever seen, and it has been proven to be higher.

So I'm going to ban now uncomputable functions from the thread, alright. *adds rule*

But still, my point stands, by the rate this thing grows you may get a bigger number with BB(g_2) from graham's sequence than with a double nest like BB(BB(9)). For instance, BB(BB(2))=13, BB(BB(BB(2))) = BB(13).

- Vytron
**Posts:**429**Joined:**Mon Oct 19, 2009 10:11 am UTC**Location:**The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P

### Re: Your number is, in fact, not bigger!

mike-l wrote:From here I think you actually underestimate a bit.

Oh yeah, this is because at some point, for some reason, I wrongly believed that ε_0^ε_0 = ω^{ε_0+1}, making it much harder for myself to reach higher ordinals, (was forcing myself to reach {ε_0*ω}^{ε_0*ω}^{ε_0*ω}... to get to ε_1, and getting to ε_{ε_...{ε_0*ω}*ω}*ω} for ζ_0 when this wasn't necessary.) And it was the reason I had plenty of extension ideas, like {0}~{0}~{0}~{0} to mean cascading ordinal multiplication before {0,0}, when it wasn't necessary at all to reach the ordinals I wanted to reach (because I wasn't realizing I was already getting to them).

My only advice to Daggoth is to stop carving carefully every tree of his forest and just create a bunch of more forests.

I had this function back in August 2014:

n[+] = n+1

n[0] = n[+]n...n[+] times...n[+]n

n[1] = n[0]n...n[0] times...n[0]n

n[d] = n[d-1]n...n[d-1] times...n[0]n

n[0,0] = n[n]

n[0,d] = n[0,n[0,n...n[0,d-1] times...[0,n[0,d-1]...]]

n[d,0] = n[d-1,n[0,3]]

n[d,d] = n[d,n[d,n...n[d,d-1] times...[d-1,n[d,d-1]...]]

n[0,0,0] = n[n,n]

n[0,0,d] = n[0,0,n[0,0,n...n[0,0,d-1] times...[0,0,n[0,0,d-1]...]]

n[0,d,0] = n[0,d-1,n[0,d-1,3]]

n[0,d,d] = n[0,d-1,n[0,d-1,n...n[0,d-1,n] times...[0,d-1,n[d-1,n]...]]

n[d,0,0] = n[d-1,n[d-1,3,0],0]

n[d,0,d] = n[d,0,n[d,0,n...n[d,0,d-1] times...[d,0,n[d,0,d-1]...]]

n[d,d,0] = n[d,d-1,n[d,d-1,3]]

n[d,d,d] = n[d,d,n[d,d,n...n[d,d,d-1] times...[d,d-1,n[d,d,d-1]...]]

n[0,0,0,0] = n[n,n,n]

n[x,x,x,d] = n[x,x,x,n[x,x,x,n...n[x,x,x,d-1] times...[x,x,x,n[x,x,x,d-1]...]]

n[x,x,d,0] = n[x,x,d-1,n[x,x,d-1,3]]

n[x,d,0,0] = n[x,d-1,n[x,d-1,3,0],0]

n[d,0,0,0] = n[d-1,n[d-1,3,0,0],0,0]

n[0,0,...d times...0,0] = n[n,n,......n,n]

n[,0] = n[0,0,...n[n,n,n] times...0,0]

It's much weaker than what I have now (goes up to f_w^w instead of e_0).

But apparently, I could have plugged the n[0-0-0-0] nested concatenations thing to it since then and reach the same values that I'm reaching now.

So on the long run I don't think all the time spent working on the details of the specific nuances of the base progression is worth it.

### Re: Your number is, in fact, not bigger!

Hey guys. It's been some time since I've posted here, and wow it's really exploded. It looks like the numbers have gotten a lot bigger too. I remember Vytron not wanting the numbers to grow too fast, but I guess it's been a couple months.

I would be happy to analyze the latest notations, if people would post the most updated versions.

I'm looking at WarDaft's notation, and I don't quite understand it - for example with

(X,0,0)-- where X contains no non-zero terms = <(X,>0<)>

I'm not quite sure what the <>'s are for.

I'm not really following Vytron's last post either, where he says:

n,1 = n,[0-0-0-0] for n -s

n,1 = φ(1@ω) so that k,1 = φ(1@k)

why are there two different definitions of n,1? And how can n,1 = φ(1@ω) if k,1 = φ(1@k)? Are n and k not variables?

Also, I'm confused by the comment "(was forcing myself to reach {ε_0*ω}^{ε_0*ω}^{ε_0*ω}... to get to ε_1, and getting to ε_{ε_...{ε_0*ω}*ω}*ω} for ζ_0 when this wasn't necessary.)" But {ε_0*ω}^{ε_0*ω}^{ε_0*ω}... = ε_1, so if you reach one you reach the other. Similarly for ε_{ε_...{ε_0*ω}*ω}*ω} = ζ_0.

I would be happy to analyze the latest notations, if people would post the most updated versions.

I'm looking at WarDaft's notation, and I don't quite understand it - for example with

(X,0,0)-- where X contains no non-zero terms = <(X,>0<)>

I'm not quite sure what the <>'s are for.

I'm not really following Vytron's last post either, where he says:

n,1 = n,[0-0-0-0] for n -s

n,1 = φ(1@ω) so that k,1 = φ(1@k)

why are there two different definitions of n,1? And how can n,1 = φ(1@ω) if k,1 = φ(1@k)? Are n and k not variables?

Also, I'm confused by the comment "(was forcing myself to reach {ε_0*ω}^{ε_0*ω}^{ε_0*ω}... to get to ε_1, and getting to ε_{ε_...{ε_0*ω}*ω}*ω} for ζ_0 when this wasn't necessary.)" But {ε_0*ω}^{ε_0*ω}^{ε_0*ω}... = ε_1, so if you reach one you reach the other. Similarly for ε_{ε_...{ε_0*ω}*ω}*ω} = ζ_0.

### Re: Your number is, in fact, not bigger!

< > is a repetition operator, so I don't have to specify with a function or keep typing "(0,0,..0) for n 0's"

5|(0,0) uses the rule (X,0,0)-- = <(X,>0<)> , X is an empty array, so <(X,> is really just <(>, so it simplifies to 5|(0,0) = 5+1|<(>0<)> = 6|(((((0)))))

5|(0,0) uses the rule (X,0,0)-- = <(X,>0<)> , X is an empty array, so <(X,> is really just <(>, so it simplifies to 5|(0,0) = 5+1|<(>0<)> = 6|(((((0)))))

All Shadow priest spells that deal Fire damage now appear green.

Big freaky cereal boxes of death.

### Re: Your number is, in fact, not bigger!

Thanks, that clears things up.

There seems to be a problem with the rule:

(0[b:]0,X)-- = <(0[>b--<:]0,X)>

That would seem to create more complicated prefixes to :, for example

(0[1:]0)-- = (0[(0[(0[(0[0:]0):]0):]0):]0)

There seems to be a problem with the rule:

(0[b:]0,X)-- = <(0[>b--<:]0,X)>

That would seem to create more complicated prefixes to :, for example

(0[1:]0)-- = (0[(0[(0[(0[0:]0):]0):]0):]0)

### Re: Your number is, in fact, not bigger!

Ah, yes, that one should read (0[b:]0,X)-- = <(>0<[b--:]0,X)>, I moved the positional term from after the square braces to the first place in them, and that line of definition got messed up in the update. I'll double check the rest.

The updated rules.

n|0 = n

n|X = n+1|X--

(b)-- = b

(X,0,0)-- where X contains no non-zero terms = <(X,>0<)>

(X,0,b,Y)-- where X contains no non-zero terms = <(X,>0<,b--,Y)>

(a,X,0,b,Y)-- where X contains no non-zero terms, and b does not contain square brackets = <(a--,X,>((a--,X,0,b,Y))<,b--,Y)>

(a,b,X)-- = <(>((a--,b,X))<,b--,X)>

(X,0--) = (X) .. that is, if we attempt to decrement a 0 at the end of an array, we simply chop it off the array.

Terms containing [:] are considered non-zero.

(0[:]0,X) = (<0,>X)

(a,0[:]0,X)-- = (<0,>((a--,0[:]0,X)),X)

(a,b[:]0,X)-- = (<0,>((a--,b[:]0,X)),b--[:]0,X)

(0[b:]0,X)-- = <(0[>b--<:]0,X)>

(a,0[b:]0,X)-- = <(>((a--,0[b:]0,X))<[b--:]0,X)>

(a[b:]0,X)-- = <(>(a--[b:]0,X)<[b--:]0,a--[b:]0,X)>

(a,b[c:]0,X)-- = <(>((a--,b[c:]0,X))<[c--:]0,b--[c:]0,X)>

(0[0,a:]0,X)-- = <(0[>0<,a--:]0,X)>

(a,0[0,b:]0,X) = <(0[>((a--,0[0,b:]0,X))<,b--:]0,X)>

The updated rules.

n|0 = n

n|X = n+1|X--

(b)-- = b

(X,0,0)-- where X contains no non-zero terms = <(X,>0<)>

(X,0,b,Y)-- where X contains no non-zero terms = <(X,>0<,b--,Y)>

(a,X,0,b,Y)-- where X contains no non-zero terms, and b does not contain square brackets = <(a--,X,>((a--,X,0,b,Y))<,b--,Y)>

(a,b,X)-- = <(>((a--,b,X))<,b--,X)>

(X,0--) = (X) .. that is, if we attempt to decrement a 0 at the end of an array, we simply chop it off the array.

Terms containing [:] are considered non-zero.

(0[:]0,X) = (<0,>X)

(a,0[:]0,X)-- = (<0,>((a--,0[:]0,X)),X)

(a,b[:]0,X)-- = (<0,>((a--,b[:]0,X)),b--[:]0,X)

(0[b:]0,X)-- = <(0[>b--<:]0,X)>

(a,0[b:]0,X)-- = <(>((a--,0[b:]0,X))<[b--:]0,X)>

(a[b:]0,X)-- = <(>(a--[b:]0,X)<[b--:]0,a--[b:]0,X)>

(a,b[c:]0,X)-- = <(>((a--,b[c:]0,X))<[c--:]0,b--[c:]0,X)>

(0[0,a:]0,X)-- = <(0[>0<,a--:]0,X)>

(a,0[0,b:]0,X) = <(0[>((a--,0[0,b:]0,X))<,b--:]0,X)>

All Shadow priest spells that deal Fire damage now appear green.

Big freaky cereal boxes of death.

### Re: Your number is, in fact, not bigger!

Here is GFDQM: The Final cut

Red parts are the ones that get deleted or reduced

Rule 1: Base cases:

¿0?n = n↑

¿X?n = ¿X-1?

Rule 2: Base separator ¡0!

2.1: ¿0¡0!0?n = ¿n?n

2.2: ¿...x¡0!y¡0!z?n where all arguments = 0 Reduces to ¿...x¡0!y?

2.3: ¿...x¡0!0¡0!0..¡0!0¡0!0?n where X isnt 0. Reduces to ¿...x-1¡0!n¡0!n...¡0!n¡0!n?

2.4: ¿...x¡0!y¡0!z?n where Z isnt 0 reduces to ¿...x¡0!y¡0!z-1?

Rule 3: A single Nth separator,

3.1 ¿0¡X!0?n = ¿Y¡X-1!Y....Y¡X-1!Y?

3.2 ¿Q¡X!0?n = ¿Q-1¡X!Y....Y¡X-1!Y?

3.3 ¿Q¡X!P?n = ¿Q¡X!P-1?

Rule 4: Chains.

4.1: The number most to the right isnt 0

¿#¡*!...Q¡X!P?n = ¿#¡*!...Q¡X!P-1?

4.2: All instances of the same separator, all 0's

¿0¡Q!0....0¡Q!0?n = ¿X...¡Q!X?

4.3: All instances of the same separator, the rightmost number is 0 but some or all of the others aren't.

From right to left, P is the first number that isnt 0. ¿0¡Q!...P...¡Q!0¡Q!0...?n = ¿0¡Q!...P-1...¡Q!X¡Q!X?

4.4: Mixed Separators.

Locate the lowest ranking of separator. Call this ¡L! Its surrounding values are P and Q Adjacent separators of the same rank are called a group.

4.4.1: If there is only one ¡L! ¿#¡*!#¡*!...P¡L!Q...#¡*!#¡*!?n then

4.4.1.1: ¿#¡*!#¡*!...0¡L!0...#¡*!#¡*!?n = ¿#¡*!#¡*!...X...#¡*!#¡*!?

IF ¡L! is ¡0! reduce according to rule 2

4.4.1.2: ¿#¡*!#¡*!...Q¡L!0...#¡*!#¡*!?n = ¿#¡*!#¡*!...Q-1¡L!X...#¡*!#¡*!?

4.4.1.3: ¿#¡*!#¡*!...Q¡L!P...#¡*!#¡*!?n = ¿#¡*!#¡*!...Q¡L!P-1...#¡*!#¡*!?

Adjacent separators of the same rank are called a group.

4.4.2: A single group of ¡L! (the lowest ranking separator).

¿#¡*!#¡*!...P¡L!Q¡L!R...#¡*!#¡*!?n

Reduce this group in the same way as 4.1, 4.2 and 4.3. the rest of the chain is untouched untill this group is reduced to 0

4.4.3: Multiple groups of ¡L!.

¿#¡*!#¡*!...P¡L!Q¡L!R...#¡*!#¡*!S¡L!T¡L!U...?n

Reduce the rightmost group first. regardless of the amount of ¡L!'s in that group

Rule 5: Nesting

All Nested separators rank above all ¡X!

Comparing two nested separators:

¡0...¡0!...0! with n nestings outranks ¡0...¡X!...0! with n-1 nestings

¡0¡R!0! outranks ¡x¡R-1!X!

¡R¡Q!0! outranks ¡R-1¡Q!X!

¡P¡Q!R! outranks ¡P¡Q!R-1!

As with rules 1-4, the lowest ranking group of separators gets reduced first.

5.1 ¿P¡#!X?n = ¿P¡#!X reduced?

5.2 ¿Q¡#!X?n = ¿Q reduced¡#!X?

5.3 ¿0¡#!0?n = ¿X¡# reduced!X?

5.4 if any of these three rules above would reduce 0¡0!0, it reduces to n

Rule 6:

Double Separator. Ranks above all nesteds. Rank according to reductions. If a Separator could eventually reduce to another, it ranks above it..

6.1 ¿0¡0!¡0!0?n = ¿X¡X...X...X!X?

6.2 ¿P¡*!¡*!S?n = ¿P¡Q!¡R!S-1?

6.3 ¿P¡*!¡*!0?n = ¿P-1¡Q!¡R!X?

6.4 ¿0¡%!¡*!0?n = ¿X¡X...¡%!¡@n!...!X!X?

6.5 ¿0¡*!¡0!0?n = ¿X¡X...¡#!¡@n!...!X!X?

Triple and quadruple separators

¿0¡0!¡0!¡0!0?n = ¿X¡X...¡@!¡@!...X!X?

¿0¡0!¡0!¡0!¡0!0?n = ¿X¡X...¡@!¡@!¡@!...X!X?

Red parts are the ones that get deleted or reduced

Rule 1: Base cases:

¿0?n = n↑

^{n}n¿X?n = ¿X-1?

^{n}n (where ¿X-n?^{3}=¿X-n?¿X-n?¿X-n?)Rule 2: Base separator ¡0!

2.1: ¿0¡0!0?n = ¿n?n

2.2: ¿...x¡0!y¡0!z?n where all arguments = 0 Reduces to ¿...x¡0!y?

^{n}n where all arguments = n2.3: ¿...x¡0!0¡0!0..¡0!0¡0!0?n where X isnt 0. Reduces to ¿...x-1¡0!n¡0!n...¡0!n¡0!n?

^{n}n2.4: ¿...x¡0!y¡0!z?n where Z isnt 0 reduces to ¿...x¡0!y¡0!z-1?

^{n}nRule 3: A single Nth separator,

3.1 ¿0¡X!0?n = ¿Y¡X-1!Y....Y¡X-1!Y?

^{n}n, (with n ¡X-1! separators) Where Y = Z¡X-2!Z...¡X-2!Z? (with n ¡X-2! separators)continue untill X=03.2 ¿Q¡X!0?n = ¿Q-1¡X!Y....Y¡X-1!Y?

^{n}n3.3 ¿Q¡X!P?n = ¿Q¡X!P-1?

^{n}nRule 4: Chains.

4.1: The number most to the right isnt 0

¿#¡*!...Q¡X!P?n = ¿#¡*!...Q¡X!P-1?

^{n}n4.2: All instances of the same separator, all 0's

¿0¡Q!0....0¡Q!0?n = ¿X...¡Q!X?

^{n}n where X = Y¡Q-1!Y....Y¡X-1!Y, (with n ¡Q-1! separators), Where Y = Z¡Q-2!Z...¡Q-2!Z? (with n ¡Q-2! separators)continue untill Q=0 the final value in this sequence is n4.3: All instances of the same separator, the rightmost number is 0 but some or all of the others aren't.

From right to left, P is the first number that isnt 0. ¿0¡Q!...P...¡Q!0¡Q!0...?n = ¿0¡Q!...P-1...¡Q!X¡Q!X?

^{n}n Where X = Y¡Q-1!Y...Y¡Q-1!Y with n ¡Q-1! separators...Where Y = Z¡Q-2!Z...¡Q-2!Z? (with n ¡Q-2! separators)4.4: Mixed Separators.

Locate the lowest ranking of separator. Call this ¡L! Its surrounding values are P and Q Adjacent separators of the same rank are called a group.

4.4.1: If there is only one ¡L! ¿#¡*!#¡*!...P¡L!Q...#¡*!#¡*!?n then

4.4.1.1: ¿#¡*!#¡*!...0¡L!0...#¡*!#¡*!?n = ¿#¡*!#¡*!...X...#¡*!#¡*!?

^{n}n Where X = Y¡L-1!Y...Y¡L-1!Y with n ¡L-1! separators where Y... ( same as with rule 3.1 )IF ¡L! is ¡0! reduce according to rule 2

4.4.1.2: ¿#¡*!#¡*!...Q¡L!0...#¡*!#¡*!?n = ¿#¡*!#¡*!...Q-1¡L!X...#¡*!#¡*!?

^{n}n X is the same as in the previous rule.4.4.1.3: ¿#¡*!#¡*!...Q¡L!P...#¡*!#¡*!?n = ¿#¡*!#¡*!...Q¡L!P-1...#¡*!#¡*!?

^{n}nAdjacent separators of the same rank are called a group.

4.4.2: A single group of ¡L! (the lowest ranking separator).

¿#¡*!#¡*!...P¡L!Q¡L!R...#¡*!#¡*!?n

Reduce this group in the same way as 4.1, 4.2 and 4.3. the rest of the chain is untouched untill this group is reduced to 0

4.4.3: Multiple groups of ¡L!.

¿#¡*!#¡*!...P¡L!Q¡L!R...#¡*!#¡*!S¡L!T¡L!U...?n

Reduce the rightmost group first. regardless of the amount of ¡L!'s in that group

Rule 5: Nesting

All Nested separators rank above all ¡X!

Comparing two nested separators:

¡0...¡0!...0! with n nestings outranks ¡0...¡X!...0! with n-1 nestings

¡0¡R!0! outranks ¡x¡R-1!X!

¡R¡Q!0! outranks ¡R-1¡Q!X!

¡P¡Q!R! outranks ¡P¡Q!R-1!

As with rules 1-4, the lowest ranking group of separators gets reduced first.

5.1 ¿P¡#!X?n = ¿P¡#!X reduced?

^{n}n5.2 ¿Q¡#!X?n = ¿Q reduced¡#!X?

^{n}n5.3 ¿0¡#!0?n = ¿X¡# reduced!X?

^{n}n5.4 if any of these three rules above would reduce 0¡0!0, it reduces to n

Rule 6:

Double Separator. Ranks above all nesteds. Rank according to reductions. If a Separator could eventually reduce to another, it ranks above it..

6.1 ¿0¡0!¡0!0?n = ¿X¡X...X...X!X?

^{n}n Where X = ¡Y...Y¡n!Y...Y! where Y = ¡Z...Z¡n-1!Z...Z! so on until ¡n-n=0! reduces to , Use n separators each step of the way6.2 ¿P¡*!¡*!S?n = ¿P¡Q!¡R!S-1?

^{n}n6.3 ¿P¡*!¡*!0?n = ¿P-1¡Q!¡R!X?

^{n}n Where X = Y¡#!Y... with n ¡#! separators where # is a separator immediately inferior in rank, Where Y is Z¡#-1!Z... with n ?¡#-1! Separators where #-1 is immediately inferior to #. Continue until # is the base separator6.4 ¿0¡%!¡*!0?n = ¿X¡X...¡%!¡@n!...!X!X?

^{n}n Where @n =Y¡Y...¡%!¡@n-1!Y...!Y ..so on untill @n-n = Z¡Z...¡%!¡#!...Z!Z Each set of ellipses hides n nestings. X Y Z work as in 6.36.5 ¿0¡*!¡0!0?n = ¿X¡X...¡#!¡@n!...!X!X?

^{n}n @ Works the same as in 6.4Triple and quadruple separators

¿0¡0!¡0!¡0!0?n = ¿X¡X...¡@!¡@!...X!X?

^{n}n Here @ Works the same as in 6.4 taking * to be n¿0¡0!¡0!¡0!¡0!0?n = ¿X¡X...¡@!¡@!¡@!...X!X?

^{n}n Here @ Works the same as in 6.4 taking * to be n### Re: Your number is, in fact, not bigger!

WarDaft, when you have (a,0), you use the (a,b,X)-- = <(>((a--,b,X))<,b--,X)> rule right?

So eg 2|(a,0) = (( ((a--, 0)) ))? This suggests that (a,0) is (a--,0) + w and inductively w(1+a). Am I using the wrong rule here?

So eg 2|(a,0) = (( ((a--, 0)) ))? This suggests that (a,0) is (a--,0) + w and inductively w(1+a). Am I using the wrong rule here?

addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.

### Re: Your number is, in fact, not bigger!

Hmm, that actually does look like the appropriate rule. It would seem I am missing a base case, because that was not the rule I was applying in my head.

Let me think... this might not actually be a problem in the long run though it would change the nature of the example proof.

(0,1) would be the first fixed point of a -> ω(1+a) IE ω^ω

(1,1) would then be <(>((0,1))<,0)>

((0,1),0) reduces to <((>0<,0)><)> = ω^n + ω thus it is still ω^ω --- (0,1) is a fixed point though, so technically shouldn't be any bigger.

(((0,1)),0) = <(>((0,1),0)<)> which properly functions as ω^ω+ω in the Hardy hierarchy, if a bit strange in how it gets there.

(1,1) reduces to <(>((0,1))<,0)>, we'll take n = 3 here, so ((( ((0,1)) ,0),0),0)

((( ((0,1)) ,0),0),0)--

= ((( ((( ((0,1)) ,0),0)--,0) )))

= ((( ( ((( (( ((0,1)) ,0)-- ,0) ))),0) )))

= ((( ( ((( ( ((( ( ((0,1))-- ,0) ))),0) ))),0) )))

= ((( ( ((( ( ((( ( (0,1) ,0) ))),0) ))),0) )))

replacing (((n))) with n+3...

= ((((0,1),0)+3,0)+3,0)+3

So it doesn't look like there will be any problem points, (b) still satisfactorily acts like b+1 to continue past fixed points, even if they weren't the points I had originally planned. It's also not as tidy as I had wanted. It quickly matches up again with my original intent by (0,0,0,0)

Let me think... this might not actually be a problem in the long run though it would change the nature of the example proof.

(0,1) would be the first fixed point of a -> ω(1+a) IE ω^ω

(1,1) would then be <(>((0,1))<,0)>

((0,1),0) reduces to <((>0<,0)><)> = ω^n + ω thus it is still ω^ω --- (0,1) is a fixed point though, so technically shouldn't be any bigger.

(((0,1)),0) = <(>((0,1),0)<)> which properly functions as ω^ω+ω in the Hardy hierarchy, if a bit strange in how it gets there.

(1,1) reduces to <(>((0,1))<,0)>, we'll take n = 3 here, so ((( ((0,1)) ,0),0),0)

((( ((0,1)) ,0),0),0)--

= ((( ((( ((0,1)) ,0),0)--,0) )))

= ((( ( ((( (( ((0,1)) ,0)-- ,0) ))),0) )))

= ((( ( ((( ( ((( ( ((0,1))-- ,0) ))),0) ))),0) )))

= ((( ( ((( ( ((( ( (0,1) ,0) ))),0) ))),0) )))

replacing (((n))) with n+3...

= ((((0,1),0)+3,0)+3,0)+3

So it doesn't look like there will be any problem points, (b) still satisfactorily acts like b+1 to continue past fixed points, even if they weren't the points I had originally planned. It's also not as tidy as I had wanted. It quickly matches up again with my original intent by (0,0,0,0)

All Shadow priest spells that deal Fire damage now appear green.

Big freaky cereal boxes of death.

### Re: Your number is, in fact, not bigger!

WarDaft wrote:Ah, yes, that one should read (0[b:]0,X)-- = <(>0<[b--:]0,X)>, I moved the positional term from after the square braces to the first place in them, and that line of definition got messed up in the update. I'll double check the rest.

The updated rules.

n|0 = n

n|X = n+1|X--

(b)-- = b

(X,0,0)-- where X contains no non-zero terms = <(X,>0<)>

(X,0,b,Y)-- where X contains no non-zero terms = <(X,>0<,b--,Y)>

(a,X,0,b,Y)-- where X contains no non-zero terms, and b does not contain square brackets = <(a--,X,>((a--,X,0,b,Y))<,b--,Y)>

(a,b,X)-- = <(>((a--,b,X))<,b--,X)>

(X,0--) = (X) .. that is, if we attempt to decrement a 0 at the end of an array, we simply chop it off the array.

Terms containing [:] are considered non-zero.

(0[:]0,X) = (<0,>X)

(a,0[:]0,X)-- = (<0,>((a--,0[:]0,X)),X)

(a,b[:]0,X)-- = (<0,>((a--,b[:]0,X)),b--[:]0,X)

(0[b:]0,X)-- = <(0[>b--<:]0,X)>

(a,0[b:]0,X)-- = <(>((a--,0[b:]0,X))<[b--:]0,X)>

(a[b:]0,X)-- = <(>(a--[b:]0,X)<[b--:]0,a--[b:]0,X)>

(a,b[c:]0,X)-- = <(>((a--,b[c:]0,X))<[c--:]0,b--[c:]0,X)>

(0[0,a:]0,X)-- = <(0[>0<,a--:]0,X)>

(a,0[0,b:]0,X) = <(0[>((a--,0[0,b:]0,X))<,b--:]0,X)>

These seem to be the old rules.

### Re: Your number is, in fact, not bigger!

Wow, I honestly don't know how I managed to fail at copy and pasting.

n|0 = n

n|X = n+1|X--

(b)-- = b

(X,0,0)-- where X contains no non-zero terms = <(X,>0<)>

(X,0,b,Y)-- where X contains no non-zero terms = <(X,>0<,b--,Y)>

(a,X,0,b,Y)-- where X contains no non-zero terms, and b does not contain square brackets = <(a--,X,>((a--,X,0,b,Y))<,b--,Y)>

(a,b,X)-- = <(>((a--,b,X))<,b--,X)>

(X,0--) = (X) .. that is, if we attempt to decrement a 0 at the end of an array, we simply chop it off the array.

Terms containing [:] are considered non-zero.

(0[:]0,X) = (<0,>X)

(a,0[:]0,X)-- = (<0,>((a--,0[:]0,X)),X)

(a,b[:]0,X)-- = (<0,>((a--,b[:]0,X)),b--[:]0,X)

(0[b:]0,X)-- = <(>0<[b--:]0,X)>

(a,0[b:]0,X)-- = <(>((a--,0[b:]0,X))<[b--:]0,X)>

(a[b:]0,X)-- = <(>(a--[b:]0,X)<[b--:]0,a--[b:]0,X)>

(a,b[c:]0,X)-- = <(>((a--,b[c:]0,X))<[c--:]0,b--[c:]0,X)>

(0[0,a:]0,X)-- = <(0[>0<,a--:]0,X)>

(a,0[0,b:]0,X) = <(0[>((a--,0[0,b:]0,X))<,b--:]0,X)>

I'm not going to claim this is a perfect version, because apparently I'm just not seeing what I've typed, merely what I expect to be there. Hopefully it is by now though.

n|0 = n

n|X = n+1|X--

(b)-- = b

(X,0,0)-- where X contains no non-zero terms = <(X,>0<)>

(X,0,b,Y)-- where X contains no non-zero terms = <(X,>0<,b--,Y)>

(a,X,0,b,Y)-- where X contains no non-zero terms, and b does not contain square brackets = <(a--,X,>((a--,X,0,b,Y))<,b--,Y)>

(a,b,X)-- = <(>((a--,b,X))<,b--,X)>

(X,0--) = (X) .. that is, if we attempt to decrement a 0 at the end of an array, we simply chop it off the array.

Terms containing [:] are considered non-zero.

(0[:]0,X) = (<0,>X)

(a,0[:]0,X)-- = (<0,>((a--,0[:]0,X)),X)

(a,b[:]0,X)-- = (<0,>((a--,b[:]0,X)),b--[:]0,X)

(0[b:]0,X)-- = <(>0<[b--:]0,X)>

(a,0[b:]0,X)-- = <(>((a--,0[b:]0,X))<[b--:]0,X)>

(a[b:]0,X)-- = <(>(a--[b:]0,X)<[b--:]0,a--[b:]0,X)>

(a,b[c:]0,X)-- = <(>((a--,b[c:]0,X))<[c--:]0,b--[c:]0,X)>

(0[0,a:]0,X)-- = <(0[>0<,a--:]0,X)>

(a,0[0,b:]0,X) = <(0[>((a--,0[0,b:]0,X))<,b--:]0,X)>

I'm not going to claim this is a perfect version, because apparently I'm just not seeing what I've typed, merely what I expect to be there. Hopefully it is by now though.

All Shadow priest spells that deal Fire damage now appear green.

Big freaky cereal boxes of death.

### Re: Your number is, in fact, not bigger!

is <> shorthand for what we(me+vytron)have been doing with ellipses?, if so may i borrow it

### Re: Your number is, in fact, not bigger!

Yeah. It's a good way of denoting things. I also quite like the -- as short hand for 'reduce this thing'

So eg my base reduction rules that Vytron and I have been using look something like:

[X,b,Y,a]-- = [X,b,<Y--, a>] when b>=a, all terms in Y are < a.

[X,b,a+1]-- = [X,b, a--, a] when b> a

Very compact and explicit.

So eg my base reduction rules that Vytron and I have been using look something like:

[X,b,Y,a]-- = [X,b,<Y--, a>] when b>=a, all terms in Y are < a.

[X,b,a+1]-- = [X,b, a--, a] when b> a

Very compact and explicit.

Last edited by mike-l on Wed Mar 04, 2015 5:04 pm UTC, edited 1 time in total.

addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.

### Re: Your number is, in fact, not bigger!

Ok, i'll give it a try to rewrite GFDQM with those symbols later this week, it reads weird to me, but if it means someone being more able to help me evaluate my notation its a good plan.

- Vytron
**Posts:**429**Joined:**Mon Oct 19, 2009 10:11 am UTC**Location:**The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P

### Re: Your number is, in fact, not bigger!

Welcome back Deedlit! I've been reading your articles at Googology wiki. By my understanding of it you could easily blow us all away but are holding back, eh?

You should probably stop hanging out with me (...because, most of the misunderstandings of my previous notations were had because of my complete failures at copy and pasting, that you have it now means it's probably contagious. Beware.)

n,[0-0-0-0] is my notation, φ(1@ω) is the FGH.

It works like in the base nesting of the FGH where if you have g(n) = f_n(n) then you have f_ω(n). So n,1 translates to φ(1,0,0,0) in the FGH for n 0s.

I'm still, utterly confused by this!

My confusion is in two parts:

1-

There's these three ordinal series:

ε_0, {ε_0*ω}, {ε_0*ω}*ω, {{ε_0*ω}*ω}*ω...

ε_0+1, ω^{ε_0+1}, ω^ω^{ε_0+1}, ω^ω^ω^{ε_0+1}...

ε_0, ε_0^ε_0, ε_0^ε_0^ε_0, ε_0^ε_0^ε_0^ε_0...

And it seem ε_1 is the limit of all of them?

Does that mean they're equivalent? Like {ε_0*ω}=ω^{ε_0+1}=ε_0^ε_0 or something? How can multiplying by ω be equal to exponentiation by ε_0? Or, if the former suffices for ε_1 why bothering going further?

2-

WarDaft says this doesn't matter. He says my concatenated notation is strong enough to reach what I want at the limit of the -s.

I take it to mean this:

There's a series of nesting, say:

n[0-0], n[[0-0]-0], n[[[0-0]-0]-0], n[[[[0-0]-0]-0]-0]

And:

n[0-0-0] enumerates them.

n[0-0-0] belongs to a value φ(x), where x is a 1 followed by a ,0 or more, if this nesting is strong enough.

If the nesting is not strong enough, then n[0-0-0] belongs to a value φ(x), where x is n followed by less ,0s than in the above case.

This means I just need an extra -0 in the box to reach the value that the "not strong enough notation reached".

I.e. if there's a series of nestings:

n[0-0-0], n[0-0-n[0-0-0]], n[0-0-[0-0-[0-0-0]]]...

Enumerated by:

n[0-x-0]

A series of nestings:

n[0-0-0], n[0-[0-0-0]-0], n[0-[0-[0-0-0]-0]-0]...

Enumerated by

n[x-0-0]

And a series of nestings:

n[0-0-0], n[[0-0-0]-0-0], n[[[0-0-0]-0-0]-0-0]...

Enumerated by:

n[0-0-0-0]

Then n[0-0-0-0] belongs to a value φ(x), where x is a 1 followed by a ,0 or more, that n[0-0-0] would have reached if it was "strong enough".

This means that if x is weak, I still reach n,1 = n[0-0-0-0] for y -s = φ(1,0,0,0) for n ,0s, at y=n+2, or at y=n*2, or at y=n^2... Or I just need to make y = n[0-0-0-0] for n-1 0s, or something, and that'll do.

Or I can make the base stronger, n[0-0-0] can enumerate instead:

n[0,0-0], n[[0,0-0]-0], n[[0,0-0]-0], n[[[0,0-0]-0]-0]

Or

n[{0}-0], n[[{0}-0]-0], n[[{0}-0]-0], n[[[{0}-0]-0]-0]

Or

n[{0,0}-0], n[[{0,0}-0]-0], n[[{0,0}-0]-0], n[[[{0,0}-0]-0]-0]

Or

n[{{0}}-0], n[[{{0}}-0]-0], n[[{{0}}-0]-0], n[[[{{0}}-0]-0]-0]

Or

n[{...{0}...}-0], n[[{...{0}...}-0]-0], n[[{...{0}...}-0]-0], n[[[{...{0}...}-0]-0]-0] for n {} nestings.

But then this last one is equivalent to the old weak progression:

n[0-0], n[[0-0]-0], n[[[0-0]-0]-0], n[[[[0-0]-0]-0]-0]

For one nesting more!

n[[0-0]-0], n[[[0-0]-0]-0], n[[[[0-0]-0]-0]-0], n[[[[[0-0]-0]-0]-0]-0]

So I can make the notation "much stronger" by nesting n+1 times instead of n times.

This can't be right, so it means the enumeration by n[0-0-0-0] with n -s is strong enough to reach φ(1,0,0,0) for n ,0s. Heck, n,1 = n[0-0-0-0] for n 0s (a -0 less) would do.

If this is true, the question is why bothering making a strong base notation to begin with? The next 0- you add here is going to eat it for breakfast, so there's some added 0- that "catches up" to a stronger notation.

That was back when my dream was for my notation to ever reach Gamma_0, which seemed very hard at the time. In my mind I had a series of symbols that were meant to reach some ordinals in the way to it, and they weren't meant to overshoot because I'd waste symbols (i.e. if I had y reaching some ordinal and my notation stopping at y+o, the +o would be wasted because it was used to get there, and I couldn't use it for stronger notations). In the end I had to drop all those symbols, and I'm not even sure using them in the future is a good idea.

Well, since WarDaft's <> and --s seem to be all the rage I'll make my notation use them (I was using ~s to denote the same, with the problem that they can't reflect iterated nesting, and I needed to pile them up to reflect where the string you need to repeat started).

Note that, so that my latest rewrite is equivalent to all my other notations, I'll be using n+1 copies of things instead of n copies, but n copies would suffice.

Define:

<x> as xx...xx for n+1 copies of x

a-- as reduce a

n, = n+n+n+n for n plus signs

n,x rules for x without mixed terms:

If the rightmost element is 0 this reduces:

n,x,0 = <(>n,x<)n,x>

n,x,{0} = n,x<,0>

n,x,{0,0} = n,x,<,0>{0}

n,x,{0,0,0} = n,x,<,0>{0}{0,0}

n,x,{{0}} = n,x,{<,0>}

n,0,x = n,x<,x>

n,x rules for x with mixed terms:

n,x = n,a,b

a Ends in the rightmost, highest element of x.

n,a,b = n,a,b--

If b is empty:

n,a = n,a,b where b is the highest element of a

n,a,b = n<,a--,b>

Examples:

n,0,0,{0},{0,0}

x is a mixed set.

b is empty (the highest element is the right most one)

a is 0,0,{0} - b is {0,0}

n,0,0,{0},{0,0} = n<<,0,{0},>{0,0}>

Boxes definition:

n,[0-0] = n,<{>0<}>

n,[0-x] = Follows the rules of above

n,[0,0-0] = n,<[0->{0}<]>

n,[0,0,0-0] = n,<[0,0->{0}<]>

n,[x-0] = Follows the rules of above

n,[0-0-0]= n,<[>{0}<-0]>

n,[0-0,0-0] = n,<[0-0>-{0}<]>

n,[0-x-0] = Follows the rules of above

n,[0,0-x-0] = n,<[0->{0}<-0]>

n,[x-0-0] = Follows the rules of above

n,[0-0-0-0]= n,<[>{0}<-0-0]>

n,[0-0-0-0-0]= n,<[>{0}<-0-0-0]>

n,[0-0-0-0-0-0]= n,<[>{0}<-0-0-0-0]>

n,1 = n,[0<-0>]

Rules for when x has an integer higher than 1:

(a,0),1 enumerates the fixed points of a -> a,1

(a,0,0),1 enumerates the fixed points of a -> (a,0),1

(a,0,0,..k 0s...0,0),1 enumerates the fixed points of a -> (a,0,0..k-1 0s...,0,0),1

(a,{0}),1 enumerates the fixed points of a -> (a<,0>),1

I send:

(3,{0},{0}),1

Where (a,{0},{0}),1 enumerates the fixed points of a -> (a,{0}<,0>),1 and (a,{0},0,0,..k 0s...0,0),1 enumerates the fixed points of a -> (a,{0},0,0..k-1 0s...,0,0),1

WarDaft wrote:Wow, I honestly don't know how I managed to fail at copy and pasting.

You should probably stop hanging out with me (...because, most of the misunderstandings of my previous notations were had because of my complete failures at copy and pasting, that you have it now means it's probably contagious. Beware.)

Deedlit wrote:I'm not really following Vytron's last post either, where he says:

n,1 = n,[0-0-0-0] for n -s

n,1 = φ(1@ω) so that k,1 = φ(1@k)

n,[0-0-0-0] is my notation, φ(1@ω) is the FGH.

It works like in the base nesting of the FGH where if you have g(n) = f_n(n) then you have f_ω(n). So n,1 translates to φ(1,0,0,0) in the FGH for n 0s.

Deedlit wrote:Also, I'm confused by the comment "(was forcing myself to reach {ε_0*ω}^{ε_0*ω}^{ε_0*ω}... to get to ε_1, and getting to ε_{ε_...{ε_0*ω}*ω}*ω} for ζ_0 when this wasn't necessary.)" But {ε_0*ω}^{ε_0*ω}^{ε_0*ω}... = ε_1, so if you reach one you reach the other. Similarly for ε_{ε_...{ε_0*ω}*ω}*ω} = ζ_0.

I'm still, utterly confused by this!

My confusion is in two parts:

1-

There's these three ordinal series:

ε_0, {ε_0*ω}, {ε_0*ω}*ω, {{ε_0*ω}*ω}*ω...

ε_0+1, ω^{ε_0+1}, ω^ω^{ε_0+1}, ω^ω^ω^{ε_0+1}...

ε_0, ε_0^ε_0, ε_0^ε_0^ε_0, ε_0^ε_0^ε_0^ε_0...

And it seem ε_1 is the limit of all of them?

Does that mean they're equivalent? Like {ε_0*ω}=ω^{ε_0+1}=ε_0^ε_0 or something? How can multiplying by ω be equal to exponentiation by ε_0? Or, if the former suffices for ε_1 why bothering going further?

2-

WarDaft says this doesn't matter. He says my concatenated notation is strong enough to reach what I want at the limit of the -s.

I take it to mean this:

There's a series of nesting, say:

n[0-0], n[[0-0]-0], n[[[0-0]-0]-0], n[[[[0-0]-0]-0]-0]

And:

n[0-0-0] enumerates them.

n[0-0-0] belongs to a value φ(x), where x is a 1 followed by a ,0 or more, if this nesting is strong enough.

If the nesting is not strong enough, then n[0-0-0] belongs to a value φ(x), where x is n followed by less ,0s than in the above case.

This means I just need an extra -0 in the box to reach the value that the "not strong enough notation reached".

I.e. if there's a series of nestings:

n[0-0-0], n[0-0-n[0-0-0]], n[0-0-[0-0-[0-0-0]]]...

Enumerated by:

n[0-x-0]

A series of nestings:

n[0-0-0], n[0-[0-0-0]-0], n[0-[0-[0-0-0]-0]-0]...

Enumerated by

n[x-0-0]

And a series of nestings:

n[0-0-0], n[[0-0-0]-0-0], n[[[0-0-0]-0-0]-0-0]...

Enumerated by:

n[0-0-0-0]

Then n[0-0-0-0] belongs to a value φ(x), where x is a 1 followed by a ,0 or more, that n[0-0-0] would have reached if it was "strong enough".

This means that if x is weak, I still reach n,1 = n[0-0-0-0] for y -s = φ(1,0,0,0) for n ,0s, at y=n+2, or at y=n*2, or at y=n^2... Or I just need to make y = n[0-0-0-0] for n-1 0s, or something, and that'll do.

Or I can make the base stronger, n[0-0-0] can enumerate instead:

n[0,0-0], n[[0,0-0]-0], n[[0,0-0]-0], n[[[0,0-0]-0]-0]

Or

n[{0}-0], n[[{0}-0]-0], n[[{0}-0]-0], n[[[{0}-0]-0]-0]

Or

n[{0,0}-0], n[[{0,0}-0]-0], n[[{0,0}-0]-0], n[[[{0,0}-0]-0]-0]

Or

n[{{0}}-0], n[[{{0}}-0]-0], n[[{{0}}-0]-0], n[[[{{0}}-0]-0]-0]

Or

n[{...{0}...}-0], n[[{...{0}...}-0]-0], n[[{...{0}...}-0]-0], n[[[{...{0}...}-0]-0]-0] for n {} nestings.

But then this last one is equivalent to the old weak progression:

n[0-0], n[[0-0]-0], n[[[0-0]-0]-0], n[[[[0-0]-0]-0]-0]

For one nesting more!

n[[0-0]-0], n[[[0-0]-0]-0], n[[[[0-0]-0]-0]-0], n[[[[[0-0]-0]-0]-0]-0]

So I can make the notation "much stronger" by nesting n+1 times instead of n times.

This can't be right, so it means the enumeration by n[0-0-0-0] with n -s is strong enough to reach φ(1,0,0,0) for n ,0s. Heck, n,1 = n[0-0-0-0] for n 0s (a -0 less) would do.

If this is true, the question is why bothering making a strong base notation to begin with? The next 0- you add here is going to eat it for breakfast, so there's some added 0- that "catches up" to a stronger notation.

Deedlit wrote:I remember Vytron not wanting the numbers to grow too fast, but I guess it's been a couple months.

That was back when my dream was for my notation to ever reach Gamma_0, which seemed very hard at the time. In my mind I had a series of symbols that were meant to reach some ordinals in the way to it, and they weren't meant to overshoot because I'd waste symbols (i.e. if I had y reaching some ordinal and my notation stopping at y+o, the +o would be wasted because it was used to get there, and I couldn't use it for stronger notations). In the end I had to drop all those symbols, and I'm not even sure using them in the future is a good idea.

Deedlit wrote:I would be happy to analyze the latest notations, if people would post the most updated versions.

Well, since WarDaft's <> and --s seem to be all the rage I'll make my notation use them (I was using ~s to denote the same, with the problem that they can't reflect iterated nesting, and I needed to pile them up to reflect where the string you need to repeat started).

Note that, so that my latest rewrite is equivalent to all my other notations, I'll be using n+1 copies of things instead of n copies, but n copies would suffice.

Define:

<x> as xx...xx for n+1 copies of x

a-- as reduce a

n, = n+n+n+n for n plus signs

n,x rules for x without mixed terms:

If the rightmost element is 0 this reduces:

n,x,0 = <(>n,x<)n,x>

n,x,{0} = n,x<,0>

n,x,{0,0} = n,x,<,0>{0}

n,x,{0,0,0} = n,x,<,0>{0}{0,0}

n,x,{{0}} = n,x,{<,0>}

n,0,x = n,x<,x>

n,x rules for x with mixed terms:

n,x = n,a,b

a Ends in the rightmost, highest element of x.

n,a,b = n,a,b--

If b is empty:

n,a = n,a,b where b is the highest element of a

n,a,b = n<,a--,b>

Examples:

n,0,0,{0},{0,0}

x is a mixed set.

b is empty (the highest element is the right most one)

a is 0,0,{0} - b is {0,0}

n,0,0,{0},{0,0} = n<<,0,{0},>{0,0}>

Boxes definition:

n,[0-0] = n,<{>0<}>

n,[0-x] = Follows the rules of above

n,[0,0-0] = n,<[0->{0}<]>

n,[0,0,0-0] = n,<[0,0->{0}<]>

n,[x-0] = Follows the rules of above

n,[0-0-0]= n,<[>{0}<-0]>

n,[0-0,0-0] = n,<[0-0>-{0}<]>

n,[0-x-0] = Follows the rules of above

n,[0,0-x-0] = n,<[0->{0}<-0]>

n,[x-0-0] = Follows the rules of above

n,[0-0-0-0]= n,<[>{0}<-0-0]>

n,[0-0-0-0-0]= n,<[>{0}<-0-0-0]>

n,[0-0-0-0-0-0]= n,<[>{0}<-0-0-0-0]>

n,1 = n,[0<-0>]

Rules for when x has an integer higher than 1:

(a,0),1 enumerates the fixed points of a -> a,1

(a,0,0),1 enumerates the fixed points of a -> (a,0),1

(a,0,0,..k 0s...0,0),1 enumerates the fixed points of a -> (a,0,0..k-1 0s...,0,0),1

(a,{0}),1 enumerates the fixed points of a -> (a<,0>),1

I send:

(3,{0},{0}),1

Where (a,{0},{0}),1 enumerates the fixed points of a -> (a,{0}<,0>),1 and (a,{0},0,0,..k 0s...0,0),1 enumerates the fixed points of a -> (a,{0},0,0..k-1 0s...,0,0),1

### Re: Your number is, in fact, not bigger!

Vytron wrote:Welcome back Deedlit! I've been reading your articles at Googology wiki. By my understanding of it you could easily blow us all away but are holding back, eh?

Thanks, Vytron! Yeah, I was following the general flow of the thread, which seemed to be "Don't overshoot the previous number by too much". I had a notation at the level of the SVO ready for when we got to that level, but we seem to have passed that now.

I know WarDaft could really dial things up as well if he wanted to.

Deedlit wrote:Also, I'm confused by the comment "(was forcing myself to reach {ε_0*ω}^{ε_0*ω}^{ε_0*ω}... to get to ε_1, and getting to ε_{ε_...{ε_0*ω}*ω}*ω} for ζ_0 when this wasn't necessary.)" But {ε_0*ω}^{ε_0*ω}^{ε_0*ω}... = ε_1, so if you reach one you reach the other. Similarly for ε_{ε_...{ε_0*ω}*ω}*ω} = ζ_0.

I'm still, utterly confused by this!

My confusion is in two parts:

1-

There's these three ordinal series:

ε_0, {ε_0*ω}, {ε_0*ω}*ω, {{ε_0*ω}*ω}*ω...

ε_0+1, ω^{ε_0+1}, ω^ω^{ε_0+1}, ω^ω^ω^{ε_0+1}...

ε_0, ε_0^ε_0, ε_0^ε_0^ε_0, ε_0^ε_0^ε_0^ε_0...

And it seem ε_1 is the limit of all of them?

Does that mean they're equivalent? Like {ε_0*ω}=ω^{ε_0+1}=ε_0^ε_0 or something? How can multiplying by ω be equal to exponentiation by ε_0? Or, if the former suffices for ε_1 why bothering going further?

Well ε_0, {ε_0*ω}, {ε_0*ω}*ω, {{ε_0*ω}*ω}*ω... doesn't go to ε_1, it goes to ε_0*ω^ω. But the series {ε_0*ω}, {ε_0*ω}^{ε_0*ω}, {ε_0*ω}^{ε_0*ω}^{ε_0*ω}... does go to ε_1. So we have three inequivalent series all going to ε_1, let's compare them.

A handy way of comparing ordinals is to write them in iterated Cantor Normal Form, which means to write them using addition and exponentiation with base ω. One can prove that you can represent any ordinal starting from 0 and the ε numbers using addition and exponentiation with base ω. So for example, ε_0^ε_0 = (ω^ε_0)^ε_0 = ω^(ε_0*ε_0) = ω^(ω^ε_0*ω^ε_0) = ω^ω^(ε_0+ε_0), which is in proper iterated Cantor Normal Form. We can then see that ε_0^ε_0 is greater than ω^ω^{ε_0+1}, but less than ω^ω^ω^{ε_0+1} = ω^ω^{ε_0*ω}. Continuing in this fashion, we can see that

ε_0 < ε_0+1 < ε_0*ω = ω^{ε_0+1} < ω^ω^{ε_0+1} < ε_0^ε_0 < {ε_0*ω}^{ε_0*ω} < ω^ω^ω^{ε_0+1} < ε_0^ε_0^ε_0 < {ε_0*ω}^{ε_0*ω}^{ε_0*ω} < ...

So we see that the three series interleave with each other all the way to infinity. Yes, it is true that the nth member of the base ω series is less than the nth member of the base ε_0 series, which is less than the nth member of the base ε_0*ω series, but that doesn't mean that the limit of the base ε_0*ω series is greater than the limit of the base ω series - in fact it cannot be greater, since if you pick some element of the base ε_0*ω series, say the nth one, I can pick a larger element of the base ω series, say the n+2nd one. So all three limits must be the same, and we call that limit ε_1.

2-

WarDaft says this doesn't matter. He says my concatenated notation is strong enough to reach what I want at the limit of the -s.

<stuff deleted>

If this is true, the question is why bothering making a strong base notation to begin with? The next 0- you add here is going to eat it for breakfast, so there's some added 0- that "catches up" to a stronger notation.

I didn't quite follow what you were saying, but it seems that you've hit upon a fundamental principle in googology - making the early parts of your notation stronger and stronger isn't going to add to the final strength, when the bulk of that strength is coming from the final part. You see this with many notations: For example, the fast-growing hierarchy starts with the function f_0(n) = n+1, not because they couldn't think of a function faster growing than n+1, but because that's all that's needed, and making a stronger function wouldn't add to the strength of the notation because it would get absolutely dominated by the recursive processes anyway. Of course, you can't make you starting function too weak - if you started with f_0(n) = n, then iterating that function does nothing, so the fast-growing hierarchy would never get off the ground. But, as long as the base fuction is strong enough to get things started, making it stronger doesn't help. We see the same thing with Bowers' array notation - initially he started with {a,b} = a+b, then changed it to {a,b} = a^b to make it more compatible with other notations ike Knuth arrow and Conway chained arrow, but going with either a+b or a^b doesn't change the strenghth of the notation. Of course, your base functions are much stronger than a^b, but the same principle applies - if the bulk of the recursion is coming from the final part of the notation, then you just need to make sure the early parts are "good enough", and there's no need to make them stronger and stronger.

This raises an interesting dilemna for people who define their notations in parts - naturally, you want to keep all the work you've done so far, so the natural thing to do is to build the nex part over all the work that's been done. However, by the principle I just mentioned, often one can greatly weaken and simplify the eariler parts and still maintain the final strength of the notation. So defining "by parts" makes the final notation more complicated than it needs to be, but on the other hand restarting from scratch every time may be more work than you want, and it also means dealing with multiple notations.

Aside - with enough recursion, not only can you overcome a weaker base function, but even a weaker recursive process. For example, take the Hardy hierarchy, defined by

H_0 (n) = n

H_{a+1} (n) = H_a (n+1)

for limit a, H_a (n) = H_a(n) = H_a[n] (n)

So it's the same as the fast-growing hierarchy, except at successor ordinals, rather than iterate the previous function n times, you just increment the input by 1. So the hierarchy starts off H_0 = n, H_1 = n+1, H_2 = n+2, H_3 = n+3... - it's weaksauce. At ω, when the fast-growing hierarchy has reached the level of the Ackermann function, we still have H_ω = 2n. So it seems incomparably weaker, and even more, the difference seems to get greater and greater, and there's no way for the Hardy Hierarchy to catch up.

But it does! The fast-growing hierarchy and Hardy hierarchy are related by f_a ~ H_(ω^a) and while the latter subscript seems to grow much faster than the former, we know that ω^ε_0 = ε_0, so f_ε_0 is roughly the same as H_ε_0, and more generally f_ε_a is roughly the same as H_ε_a. So as long as your notation reaches ε_0 or further, it doesn't matter too much whether you recursive process is iteration or just incrementing the input by 1.

Vytron wrote:n,x rules for x without mixed terms:

What are "mixed terms" exactly?

If the rightmost element is 0 this reduces:

n,x,0 = <(>n,x<)n,x>

n,x,{0} = n,x<,0>

n,x,{0,0} = n,x,<,0>{0}

n,x,{0,0,0} = n,x,<,0>{0}{0,0}

n,x,{{0}} = n,x,{<,0>}

n,0,x = n,x<,x>

Wait, when does this last rule apply? it seems it can apply to most of the cases where we use the other rules, so there is a conflict.

Vytron wrote:n,x rules for x with mixed terms:

n,x = n,a,b

a Ends in the rightmost, highest element of x.

n,a,b = n,a,b--

If b is empty:

n,a = n,a,b where b is the highest element of a

n,a,b = n<,a--,b>

Do you have a justification for why n<a--,b> would be of lower order than n,a,b? It seems to me that this would go infinite.

n,a,b would go to n,...,b,a--,b, and then you would keep reducing a-- until it goes away, and you are left with n,...,b,b which is of higher order than n,a,b as b is of higher order than a.

Vytron wrote:Boxes definition:

n,[0-0] = n,<{>0<}>

n,[0-x] = Follows the rules of above

n,[0,0-0] = n,<[0->{0}<]>

n,[0,0,0-0] = n,<[0,0->{0}<]>

n,[x-0] = Follows the rules of above

n,[0-0-0]= n,<[>{0}<-0]>

n,[0-0,0-0] = n,<[0-0>-{0}<]>

n,[0-x-0] = Follows the rules of above

n,[0,0-x-0] = n,<[0->{0}<-0]>

n,[x-0-0] = Follows the rules of above

n,[0-0-0-0]= n,<[>{0}<-0-0]>

n,[0-0-0-0-0]= n,<[>{0}<-0-0-0]>

n,[0-0-0-0-0-0]= n,<[>{0}<-0-0-0-0]>

n,1 = n,[0<-0>]

I'm not sure what "Follows the rules of above" refers to, as there are no previous rules dealing with dashes and boxes.

Vytron wrote:Rules for when x has an integer higher than 1:

(a,0),1 enumerates the fixed points of a -> a,1

(a,0,0),1 enumerates the fixed points of a -> (a,0),1

(a,0,0,..k 0s...0,0),1 enumerates the fixed points of a -> (a,0,0..k-1 0s...,0,0),1

(a,{0}),1 enumerates the fixed points of a -> (a<,0>),1

This is a little clumsy, as your notations are not deflined to be ordinals, so it would be better to stick with the original method of saying how the notations reduce.

There is a bit of a problem with treating your notations as ordinals, as the notations have n embedded in them, and the n reduces the notation to a single finite number. So perhaps it would be better to separate the n from the rest of the notation.

It looks like you have a,1 where a is a notation itself, in particular it can be a sequence separated by commas. But if a = 0,0, I don't think that you want a,1 to be 0,0,1, rather you want it to be a,1 where a is 0,0. So you need something to serve as parentheses to make the distinction. However, parentheses are already part of your notation, as are brackets and braces, and angle brackets are used as meta-notation, so there's a bit of a problem here.

### Re: Your number is, in fact, not bigger!

In your three series, the last two ahave the same limit of e_1. The first has a limit of e_0 * w^w, much lower.

Having the same limit does not imply that the constituents are the same, just that for any element of the first sequence, there is somewhere in the second sequence a larger element, and vice versa.

Eg, both 1,2,3,... and 1,4,9,16,... have limits of w, but any particular term in the latter is larger than the corresponding term in the former. It's simply that at any fixed number in the second is eventually surpassed in the first. Using the faster growing sequence helps a lot at that particular ordinal, but unless the relationship between the sequences grows at a much higher rate than the corresponding ordinal (eg if your fundamental sequence for w is f_e_0) then this extra growth disappears immediately.

Now the particular two sequences you ask about actually are pretty much the same though never equal. For this note that the following identities holds: a^(bc) = (a^b)^c and a^(b+c) = a^b * a^c. For any a >= w, a = 1+a, so eg e_0^w = e_0^(1+w) = e_0*e_0^w.

Edit: ninjad but pretty much same thing Deedlit said. Leaving it since I go the opposite way in comparing the sequences.

So the 2nd series goes like:

e_0 + 1

w^(e_0+1) = w^e_0*w^1 = e_0*w

w^(e_0*w) = (w^e_0)^w = e_0^w

w^(e_0^w) = w^(e_0*e_0^w) = (w^e_0)^e_0^w = e_0^e_0^w

And continues like this, so adding a w at the base is, from here, the same as adding an e_0 at the base. So this series, after the first few terms, is a tower of e_0s capped with a w. The third series just lacks the cap, but it's clear that these now interleave:

e_0^...^e_0 < e_0^...^e_0^w < e_0^...^e_0^e_0

Having the same limit does not imply that the constituents are the same, just that for any element of the first sequence, there is somewhere in the second sequence a larger element, and vice versa.

Eg, both 1,2,3,... and 1,4,9,16,... have limits of w, but any particular term in the latter is larger than the corresponding term in the former. It's simply that at any fixed number in the second is eventually surpassed in the first. Using the faster growing sequence helps a lot at that particular ordinal, but unless the relationship between the sequences grows at a much higher rate than the corresponding ordinal (eg if your fundamental sequence for w is f_e_0) then this extra growth disappears immediately.

Now the particular two sequences you ask about actually are pretty much the same though never equal. For this note that the following identities holds: a^(bc) = (a^b)^c and a^(b+c) = a^b * a^c. For any a >= w, a = 1+a, so eg e_0^w = e_0^(1+w) = e_0*e_0^w.

Edit: ninjad but pretty much same thing Deedlit said. Leaving it since I go the opposite way in comparing the sequences.

So the 2nd series goes like:

e_0 + 1

w^(e_0+1) = w^e_0*w^1 = e_0*w

w^(e_0*w) = (w^e_0)^w = e_0^w

w^(e_0^w) = w^(e_0*e_0^w) = (w^e_0)^e_0^w = e_0^e_0^w

And continues like this, so adding a w at the base is, from here, the same as adding an e_0 at the base. So this series, after the first few terms, is a tower of e_0s capped with a w. The third series just lacks the cap, but it's clear that these now interleave:

e_0^...^e_0 < e_0^...^e_0^w < e_0^...^e_0^e_0

- Vytron
**Posts:**429**Joined:**Mon Oct 19, 2009 10:11 am UTC**Location:**The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P

### Re: Your number is, in fact, not bigger!

Thanks Deedlit, it does help to know that the reason of my confusion was not knowing that a function strong enough that it produces values of the kind ε_0+1, ω^{ε_0+1}, ω^ω^{ε_0+1}, ω^ω^ω^{ε_0+1}... then it is strong enough to reach ε_1. I thought those series were all below ε_0^ε_0.

Can you still send it? Your last notation was very cool. Also, I think we're 3 extenstions beyond SVO (the extenstion for (ω+1), for (ω2) and for (ωn)) so I don't think you'd have much problems catching up with it.

Anything beyond 0s and {0}s in x. I.e. with just 0s and {0} it's easy:

If the rightmost term is a 0, iterate the function.

If it's a {0}, look for a string of 0s prior to it, drop one of the 0s, and make n copies of the remaining 0s and any {0} at the right of them.

If you don't find any 0s, the {0} explodes into n 0s.

When you have {0,0} or higher in there you have to find the string to reduce first.

When x is a value higher than the rightmost element.

Would it help if instead of recycling a and b I named them c and d for later ruling? It looks as follows:

n,x rules for x with mixed terms:

n,x = n,a,b

a Ends in the rightmost, highest element of x.

n,a,b = n,a,b--

If b is empty:

n,a = n,c,d where d is the highest element of a

n,c,d = n<,c--,d>

Let's look at a random string:

n,{0,0},0,{0},0,0,{0,0,0},{0},{0}

By the rules, x ends at {0,0,0} (the highest element)

So we can ignore x and look at:

n,x,{0},{0}

There isn't any 0 in the string, so x eats the left {0}:

n,x,{0}

So, you reduce {0} and make n copies:

n,x<,0>

n,{0,0},0,{0},0,0,{0,0,0},{0},{0} = n,{0,0},0,{0},0,0,{0,0,0},{0}<,0>

Another:

n,{0},{0},{0,0},0,{0},0,0,{0,0,0}

Here, b is empty (the highest element in the string is the highest one), so we apply the rules where:

c={0},{0},{0,0},0,{0},0,0 - d = {0,0,0}

c should decrease, so we drop a 0, then we return:

<<{0},{0},{0,0},0,{0},0>{0,0,0}>

n,{0},{0},{0,0},0,{0},0,0,{0,0,0} = n<<,{0},{0},{0,0},0,{0},0,>,{0,0,0}>

n,{0},{0},{0,0},0,{0},0,{0,0,0} = n<<,{0},{0},{0,0},0,{0},>,{0,0,0}>

n,{0},{0},{0,0},0,{0},{0,0,0} = n<<,{0},{0},{0,0}<,{0}>>,{0,0,0}>

n,{0},{0},{0,0},{0},{0},{0},{0,0,0} = n<<,{0},{0},{0,0},{0},{0}<,0>>,{0,0,0}>

n,{0},{0},{0,0},{0},{0},0,0,0,{0,0,0} = n<<,{0},{0},{0,0},{0},{0},0,0>,{0,0,0}>

n,{0},{0},{0,0},{0},{0},0,0,{0,0,0} = n<<,{0},{0},{0,0},{0},{0},0>,{0,0,0}>

n,{0},{0},{0,0},{0},{0},0,{0,0,0} = n<<,{0},{0},{0,0},{0},{0}>,{0,0,0}>

n,{0},{0},{0,0},{0},{0},{0,0,0} = n<<,{0},{0},{0,0},{0}<,0>>,{0,0,0}>

n,{0},{0},{0,0},{0},{0,0,0} = n<<,{0},{0},{0,0},<,0>>,{0,0,0}>

n,{0},{0},{0,0},{0,0,0}

In this case, c and d are themselves mixed terms, so you have to find e and f:

e = ,{0},{0} - f = {0,0}

Reduce e and make n copies:

,{0},<,0>

<,{0},<,0>,{0,0}

n,{0},{0},{0,0},{0,0,0} = n,<<,{0},<,0>,{0,0}>,{0,0,0}>

n,{0,0},{0,0,0}

Here, {0,0} is the rightmost term of a, so you get:

{0,0} = <,0>{0}

n,{0,0},{0,0,0} = n,<<<,0>{0},>{0,0,0}>

It would be helpful to see an example where you think one is caught in an infinite loop.

Also, mike summarized these rules in two lines:

[X,b,Y,a]-- = [X,b,<Y--, a>] when b>=a, all terms in Y are < a.

[X,b,a+1]-- = [X,b, a--, a] when b> a

There's no new rules here, you just treat [0-0] as something higher than any previous value, so the same rules apply because:

n,{0},{0},{0,0},{0},{0},0,0,[0-0] = n<<,{0},{0},{0,0},{0},{0},0>,[0-0]>

Making:

n,[0-0,0] = n,<{>[0-0]<}>

n,[0-0,0,0] = n,<{>[0-0,0]<}>

So x in:

[0-x]

Can follow the base progression, i.e.:

n,[0-{0},{0},{0,0},{0},{0},0,0,[0-0]] = n,<{>[0-<<,{0},{0},{0,0},{0},{0},0>,[0-0]><}>

So there can be some nesting there:

n,[0-{0}] = n,<{>[0-<,0>]<}>

n,[0-[0-{0}]] = n,<{>[0-<{>[0-<,0>]<}>]<}>

n,[0-[0-[0-{0}]]] = n,<{>[0-<{>[0-<{>[0-<,0>]<}>]<}>]<}>

And:

n,[0,0-0] = n,<[0->{0}<]>

And x in n,[x-0] follows the same progression, so n,[0-0-0] can be the limit of that, i.e. n,<[>{0}<-0]>

Ah, but if I did that then it is known that the limit of this notation is ψ(Ω^Ω^Ω^Ω). I basically gave up my notation because I saw I'd get stuck, so extensions would be useless (mike supported this view on his analysis of the notation) and just waited to see how WarDaft did it. To go beyond the SVO, Wardaft does not define his notations to be ordinals but still say the first fixed point of a -> (0[a:]0) is is ψ(Ω^Ω^Ω). For every term in his notation I have a more or less equal term, so I should also have such fixed points, and just need to enumerate them. It's clear extra ,0s should be used to enumerate fixed points to define stronger notations, instead of ,0 nesting n twice.

It is already separated, i.e. in n,x n does never appear in x (actually, before n,1 x never contains an integer higher than 0.)

There's a reason spaces are never used in my notation anywhere. In this case, I could just use spaces to separate n. If a = 0,0 then 0,0 ,1 indicates such a separation.

Or, if ()s have to be used to denote a, then I can use spaces here instead:

a,0 ,1 enumerates the fixed points of a -> a ,1

a,0,0 ,1 enumerates the fixed points of a -> a,0 ,1

a,0,0,..k 0s...0,0 ,1 enumerates the fixed points of a -> a,0,0..k-1 0s...,0,0 ,1

a,{0} ,1 enumerates the fixed points of a -> a<,0> ,1

This is because WarDaft's way is the only way I know to reach these powers, if you manage to beat WarDaft without relying on fixed points I could see how and stick with the original method of saying how the notation reduces.

Ninjaed by mike-l:

Thanks. So much ordinal power, yet it seems they can pass each other at n+1 repetitions.

Deedlit wrote:I had a notation at the level of the SVO ready for when we got to that level, but we seem to have passed that now.

Can you still send it? Your last notation was very cool. Also, I think we're 3 extenstions beyond SVO (the extenstion for (ω+1), for (ω2) and for (ωn)) so I don't think you'd have much problems catching up with it.

Vytron wrote:n,x rules for x without mixed terms:

Deedlit wrote:What are "mixed terms" exactly?

Anything beyond 0s and {0}s in x. I.e. with just 0s and {0} it's easy:

If the rightmost term is a 0, iterate the function.

If it's a {0}, look for a string of 0s prior to it, drop one of the 0s, and make n copies of the remaining 0s and any {0} at the right of them.

If you don't find any 0s, the {0} explodes into n 0s.

When you have {0,0} or higher in there you have to find the string to reduce first.

If the rightmost element is 0 this reduces:

n,x,0 = <(>n,x<)n,x>

n,x,{0} = n,x<,0>

n,x,{0,0} = n,x,<,0>{0}

n,x,{0,0,0} = n,x,<,0>{0}{0,0}

n,x,{{0}} = n,x,{<,0>}

n,0,x = n,x<,x>

Deedlit wrote:Wait, when does this last rule apply? it seems it can apply to most of the cases where we use the other rules, so there is a conflict.

When x is a value higher than the rightmost element.

Deedlit wrote:Vytron wrote:n,x rules for x with mixed terms:

n,x = n,a,b

a Ends in the rightmost, highest element of x.

n,a,b = n,a,b--

If b is empty:

n,a = n,a,b where b is the highest element of a

n,a,b = n<,a--,b>

Do you have a justification for why n<a--,b> would be of lower order than n,a,b? It seems to me that this would go infinite.

n,a,b would go to n,...,b,a--,b, and then you would keep reducing a-- until it goes away, and you are left with n,...,b,b which is of higher order than n,a,b as b is of higher order than a.

Would it help if instead of recycling a and b I named them c and d for later ruling? It looks as follows:

n,x rules for x with mixed terms:

n,x = n,a,b

a Ends in the rightmost, highest element of x.

n,a,b = n,a,b--

If b is empty:

n,a = n,c,d where d is the highest element of a

n,c,d = n<,c--,d>

Let's look at a random string:

n,{0,0},0,{0},0,0,{0,0,0},{0},{0}

By the rules, x ends at {0,0,0} (the highest element)

So we can ignore x and look at:

n,x,{0},{0}

There isn't any 0 in the string, so x eats the left {0}:

n,x,{0}

So, you reduce {0} and make n copies:

n,x<,0>

n,{0,0},0,{0},0,0,{0,0,0},{0},{0} = n,{0,0},0,{0},0,0,{0,0,0},{0}<,0>

Another:

n,{0},{0},{0,0},0,{0},0,0,{0,0,0}

Here, b is empty (the highest element in the string is the highest one), so we apply the rules where:

c={0},{0},{0,0},0,{0},0,0 - d = {0,0,0}

c should decrease, so we drop a 0, then we return:

<<{0},{0},{0,0},0,{0},0>{0,0,0}>

n,{0},{0},{0,0},0,{0},0,0,{0,0,0} = n<<,{0},{0},{0,0},0,{0},0,>,{0,0,0}>

n,{0},{0},{0,0},0,{0},0,{0,0,0} = n<<,{0},{0},{0,0},0,{0},>,{0,0,0}>

n,{0},{0},{0,0},0,{0},{0,0,0} = n<<,{0},{0},{0,0}<,{0}>>,{0,0,0}>

n,{0},{0},{0,0},{0},{0},{0},{0,0,0} = n<<,{0},{0},{0,0},{0},{0}<,0>>,{0,0,0}>

n,{0},{0},{0,0},{0},{0},0,0,0,{0,0,0} = n<<,{0},{0},{0,0},{0},{0},0,0>,{0,0,0}>

n,{0},{0},{0,0},{0},{0},0,0,{0,0,0} = n<<,{0},{0},{0,0},{0},{0},0>,{0,0,0}>

n,{0},{0},{0,0},{0},{0},0,{0,0,0} = n<<,{0},{0},{0,0},{0},{0}>,{0,0,0}>

n,{0},{0},{0,0},{0},{0},{0,0,0} = n<<,{0},{0},{0,0},{0}<,0>>,{0,0,0}>

n,{0},{0},{0,0},{0},{0,0,0} = n<<,{0},{0},{0,0},<,0>>,{0,0,0}>

n,{0},{0},{0,0},{0,0,0}

In this case, c and d are themselves mixed terms, so you have to find e and f:

e = ,{0},{0} - f = {0,0}

Reduce e and make n copies:

,{0},<,0>

<,{0},<,0>,{0,0}

n,{0},{0},{0,0},{0,0,0} = n,<<,{0},<,0>,{0,0}>,{0,0,0}>

n,{0,0},{0,0,0}

Here, {0,0} is the rightmost term of a, so you get:

{0,0} = <,0>{0}

n,{0,0},{0,0,0} = n,<<<,0>{0},>{0,0,0}>

It would be helpful to see an example where you think one is caught in an infinite loop.

Also, mike summarized these rules in two lines:

[X,b,Y,a]-- = [X,b,<Y--, a>] when b>=a, all terms in Y are < a.

[X,b,a+1]-- = [X,b, a--, a] when b> a

Deedlit wrote:Vytron wrote:Boxes definition:

n,[0-0] = n,<{>0<}>

n,[0-x] = Follows the rules of above

n,[0,0-0] = n,<[0->{0}<]>

n,[0,0,0-0] = n,<[0,0->{0}<]>

n,[x-0] = Follows the rules of above

n,[0-0-0]= n,<[>{0}<-0]>

n,[0-0,0-0] = n,<[0-0>-{0}<]>

n,[0-x-0] = Follows the rules of above

n,[0,0-x-0] = n,<[0->{0}<-0]>

n,[x-0-0] = Follows the rules of above

n,[0-0-0-0]= n,<[>{0}<-0-0]>

n,[0-0-0-0-0]= n,<[>{0}<-0-0-0]>

n,[0-0-0-0-0-0]= n,<[>{0}<-0-0-0-0]>

n,1 = n,[0<-0>]

I'm not sure what "Follows the rules of above" refers to, as there are no previous rules dealing with dashes and boxes.

There's no new rules here, you just treat [0-0] as something higher than any previous value, so the same rules apply because:

n,{0},{0},{0,0},{0},{0},0,0,[0-0] = n<<,{0},{0},{0,0},{0},{0},0>,[0-0]>

Making:

n,[0-0,0] = n,<{>[0-0]<}>

n,[0-0,0,0] = n,<{>[0-0,0]<}>

So x in:

[0-x]

Can follow the base progression, i.e.:

n,[0-{0},{0},{0,0},{0},{0},0,0,[0-0]] = n,<{>[0-<<,{0},{0},{0,0},{0},{0},0>,[0-0]><}>

So there can be some nesting there:

n,[0-{0}] = n,<{>[0-<,0>]<}>

n,[0-[0-{0}]] = n,<{>[0-<{>[0-<,0>]<}>]<}>

n,[0-[0-[0-{0}]]] = n,<{>[0-<{>[0-<{>[0-<,0>]<}>]<}>]<}>

And:

n,[0,0-0] = n,<[0->{0}<]>

And x in n,[x-0] follows the same progression, so n,[0-0-0] can be the limit of that, i.e. n,<[>{0}<-0]>

Deedlit wrote:This is a little clumsy, as your notations are not deflined to be ordinals, so it would be better to stick with the original method of saying how the notations reduce.

Ah, but if I did that then it is known that the limit of this notation is ψ(Ω^Ω^Ω^Ω). I basically gave up my notation because I saw I'd get stuck, so extensions would be useless (mike supported this view on his analysis of the notation) and just waited to see how WarDaft did it. To go beyond the SVO, Wardaft does not define his notations to be ordinals but still say the first fixed point of a -> (0[a:]0) is is ψ(Ω^Ω^Ω). For every term in his notation I have a more or less equal term, so I should also have such fixed points, and just need to enumerate them. It's clear extra ,0s should be used to enumerate fixed points to define stronger notations, instead of ,0 nesting n twice.

Deedlit wrote:There is a bit of a problem with treating your notations as ordinals, as the notations have n embedded in them, and the n reduces the notation to a single finite number. So perhaps it would be better to separate the n from the rest of the notation.

It is already separated, i.e. in n,x n does never appear in x (actually, before n,1 x never contains an integer higher than 0.)

Deedlit wrote:It looks like you have a,1 where a is a notation itself, in particular it can be a sequence separated by commas. But if a = 0,0, I don't think that you want a,1 to be 0,0,1, rather you want it to be a,1 where a is 0,0. So you need something to serve as parentheses to make the distinction. However, parentheses are already part of your notation, as are brackets and braces, and angle brackets are used as meta-notation, so there's a bit of a problem here.

There's a reason spaces are never used in my notation anywhere. In this case, I could just use spaces to separate n. If a = 0,0 then 0,0 ,1 indicates such a separation.

Or, if ()s have to be used to denote a, then I can use spaces here instead:

a,0 ,1 enumerates the fixed points of a -> a ,1

a,0,0 ,1 enumerates the fixed points of a -> a,0 ,1

a,0,0,..k 0s...0,0 ,1 enumerates the fixed points of a -> a,0,0..k-1 0s...,0,0 ,1

a,{0} ,1 enumerates the fixed points of a -> a<,0> ,1

This is because WarDaft's way is the only way I know to reach these powers, if you manage to beat WarDaft without relying on fixed points I could see how and stick with the original method of saying how the notation reduces.

Ninjaed by mike-l:

Thanks. So much ordinal power, yet it seems they can pass each other at n+1 repetitions.

### Re: Your number is, in fact, not bigger!

Vytron wrote:If it's a {0}, look for a string of 0s prior to it, drop one of the 0s, and make n copies of the remaining 0s and any {0} at the right of them.

Wait, that seems different from the rule

n,x,{0} = n,x<,0>

which is the rule that would seem to apply here.

n,0,x = n,x<,x>

Wait, when does this last rule apply? it seems it can apply to most of the cases where we use the other rules, so there is a conflict.

When x is a value higher than the rightmost element.

But isn't x either the rightmost element of n,0,x, or a string that contains the rightmost element? When is it the case that x is higher?

Vytron wrote:Let's look at a random string:

n,{0,0},0,{0},0,0,{0,0,0},{0},{0}

By the rules, x ends at {0,0,0} (the highest element)

So we can ignore x and look at:

n,x,{0},{0}

There isn't any 0 in the string, so x eats the left {0}:

n,x,{0}

Wait, the rules don't say anything about a eating part of b. It just says to take n,a,b--, so we should just reduce {0},{0} as if a did not exist.

Vytron wrote:So, you reduce {0} and make n copies:

n,x<,0>

n,{0,0},0,{0},0,0,{0,0,0},{0},{0} = n,{0,0},0,{0},0,0,{0,0,0},{0}<,0>

Another:

n,{0},{0},{0,0},0,{0},0,0,{0,0,0}

Here, b is empty (the highest element in the string is the highest one), so we apply the rules where:

c={0},{0},{0,0},0,{0},0,0 - d = {0,0,0}

c should decrease, so we drop a 0, then we return:

<<{0},{0},{0,0},0,{0},0>{0,0,0}>

n,{0},{0},{0,0},0,{0},0,0,{0,0,0} = n<<,{0},{0},{0,0},0,{0},0,>,{0,0,0}>

This is where it seems like it would go infinite.

Let's fix n at 3, and say we had 3,0,{0,0,0}. By the rule you just described above, this evaluates to

3,<<>,{0,0,0}> = 3<,{0,0,0}> = 3,{0,0,0}{0,0,0},{0,0,0}

Note that we now have more {0,0,0}'s than before. If we try to evaluate this again, we get something like

3<<,{0,0,0},{0,0,0}-->,{0,0,0}> = 3,{0,0,0},{0,0,0}--,{0,0,0},{0,0,0}--,{0,0,0},{0,0,0}--,{0,0,0},{0,0,0},{0,0,0}--,{0,0,0},{0,0,0}--,{0,0,0},{0,0,0}--,{0,0,0},{0,0,0},{0,0,0}--,{0,0,0},{0,0,0}--,{0,0,0},{0,0,0}--,{0,0,0}

where the {0,0,0}--'s are whatever {0,0,0} evaluates to. The point is that there are now more {0,0,0}'s than there were before, and there are going to keep being more and more.

[X,b,Y,a]-- = [X,b,<Y--, a>] when b>=a, all terms in Y are < a.

[X,b,a+1]-- = [X,b, a--, a] when b> a

what is a+1? So far everything is composed of 0's.

There's no new rules here, you just treat [0-0] as something higher than any previous value, so the same rules apply because:

n,{0},{0},{0,0},{0},{0},0,0,[0-0] = n<<,{0},{0},{0,0},{0},{0},0>,[0-0]>

Ah, I get it.

Ah, but if I did that then it is known that the limit of this notation is ψ(Ω^Ω^Ω^Ω). I basically gave up my notation because I saw I'd get stuck, so extensions would be useless (mike supported this view on his analysis of the notation) and just waited to see how WarDaft did it. To go beyond the SVO, Wardaft does not define his notations to be ordinals but still say the first fixed point of a -> (0[a:]0) is is ψ(Ω^Ω^Ω). For every term in his notation I have a more or less equal term, so I should also have such fixed points, and just need to enumerate them. It's clear extra ,0s should be used to enumerate fixed points to define stronger notations, instead of ,0 nesting n twice.

Actually, Wardaft's definition of his notation http://echochamber.me/viewtopic.php?f=14&t=108893&p=3749640#p3749649 does not contain any references to fixed points or ordinals at all, it is in analysis that he talks about what ordinals his notation correspond with, and that is perfectly okay. It is there that he mentions that certain ordinals are the fixed points of certain functions.

You can examine WarDaft's definition to see how he defines his original notation so that he gets fixed points. Generally speaking, if you have a increasing and continuous ordinal function f(a), then the first fixed point of f(a) will be the limit of {0, f(0), f(f(0)), f(f(f(0))),...}, if F(a) is the ath fixed point of f(x) then F(a+1) will be the limit of {F(a)+1, f(F(a)+1), f(f(F(a)+1)),...}, and if a is limit than lim F(a) = F(lim a) (so we can define F(a)[n] = F(a[n]).

There's a reason spaces are never used in my notation anywhere. In this case, I could just use spaces to separate n. If a = 0,0 then 0,0 ,1 indicates such a separation.

Unfortunately spaces are nondirectional, so it is not clear if a space means you are enclosing the part to the right or unenclosing the part to the left.

### Re: Your number is, in fact, not bigger!

GFDQM

<> Indicates "copy pase this element n times"

-- Indicates "reduce this"

() Indicates "Group this".

copy paste <things> inside () before those outside

Capital letters indicate a natural number (nonzero)

// indicates commentary

¿? is an operator it's unary it takes natural numbers to its right and resolves from right to left when found in multiples

¡#!,¡%!, ¡@! are separator. What it does is basically separate two numbers

@ Indicates a cascade. for an example see step 5

Base cases:

Step 1: ¿0?n = n↑

Step 2: ¿X?n = <(¿X--?)>n // ω+X

Simple Separators:

Step 3: ¿0¡0!0?n = <(¿n?)>n // ω2+1

Step 4: ¿X¡Y!Z>?n = <(¿X¡Y!Z--?)>n

Step 5: ¿X¡Y!0?n = <¿X--¡Y!@>n Cascade explained

@ = @

@

...Go on untill

@y-y = n<¡0!n>

This is similar to how factorial works. Reduce to n copies of the ¡n-1! separator all the way to ¡0!.

example: ¿1¡2!0?3

Step 6: ¿0¡Y!0?n = <¿@¡Y--!@>n

From here on, (X¡Y!Z)-- means reduce applying rules 1 through 6

Chains: all separators are the same. examples: 3¡0!4¡0!5 and 100¡2!25¡2!50

Step 7: ¿P¡%!Q¡%!R?n = <¿P¡%!Q¡%!R--?>n

Step 8: ¿P¡%!Q¡%!0?n = <¿P¡%!Q--¡%!@?>n

Step 9: ¿P¡%!0¡%!0?n = <¿P--¡%!@¡%!@?>n

Step 10: ¿0¡%!0¡%!0?n = <¿@¡%!@?>n

Step 11: ¿0¡%!0¡%!0¡%!0?n = <¿@¡%!@¡%!@?>n // all-zero elements with all-equal separators eats a separator and an element

Below is chains where there are different separators, ¡L! denotes the lowest ranking one.

Step 12: ¿X¡%!P¡L!Q¡%!Y?n = <¿X¡%!(P¡L!Q)--¡%!Y?>n This ignores wether Y is 0 or not

Step 13: ¿X¡%!P¡L!Q¡L!R¡%!Y?n = <¿X¡%!(P¡L!Q¡L!R)--¡%!Y?>n This ignores wether Y is 0 or not

Step 14: ¿X¡%!P¡L!Q¡L!R¡%!Y¡L!Z¡%!W?n = <¿X¡%!P¡L!Q¡L!R¡%!(Y¡L!Z)--¡%!W?>n This ignores wether W is 0 or not

The "active" part or, the one to be reduced, is the lowest ranking and rightmost group.

Nesting. The same chain-handling and cascading rules apply to all kinds of separators.

Step 15: ¿0¡0¡0!0!0?n = <¿@¡@!@?>n Begin cascade at n<¡n!n>

Example: ¿0¡0¡0!0!0?3 = <¿@¡3!@¡3!@¡3!@>

Step 16: ¿X¡%!Y?n = <¿X¡%!Y--?>n

Step 17: ¿X¡%!0?n = <¿X--¡%!@?>n

Step 18: ¿0¡%!0?n = <¿@¡%--!@?>n

Step 15b: ¿0¡0¡0¡0!0!0!0?n = <¿@¡@¡@!@!@?>n

example: ¿0¡0¡0¡0!0!0!0?3= <¿@¡@¡3!@¡3!@¡3!@!@>n

All-zero elements (not separators) on the nested separators eat a level of nesting.

Step 19: ¿0¡0!¡0!0?n = <@¡@!@>n //

Step 20: ¿0¡#!¡%!0?n = <@¡#!¡(%)--!@>n % Can be nested

Step 21: ¿0¡%!¡0!0?n = <@¡(%)--!¡@!@>n % Can be nested

Step 19b: ¿0¡0!¡0!¡0!0?n = <@¡@!¡@!@>n

Edit: Hopefully made the wording clearer

<> Indicates "copy pase this element n times"

-- Indicates "reduce this"

() Indicates "Group this".

copy paste <things> inside () before those outside

Capital letters indicate a natural number (nonzero)

// indicates commentary

¿? is an operator it's unary it takes natural numbers to its right and resolves from right to left when found in multiples

¡#!,¡%!, ¡@! are separator. What it does is basically separate two numbers

@ Indicates a cascade. for an example see step 5

Base cases:

Step 1: ¿0?n = n↑

^{n}n // ωStep 2: ¿X?n = <(¿X--?)>n // ω+X

Simple Separators:

Step 3: ¿0¡0!0?n = <(¿n?)>n // ω2+1

Step 4: ¿X¡Y!Z>?n = <(¿X¡Y!Z--?)>n

Step 5: ¿X¡Y!0?n = <¿X--¡Y!@>n Cascade explained

@ = @

_{y-1}<¡Y--!@_{y-1}>@

_{y-1}= @_{y-2}<¡(Y--)--!@_{y-2}>...Go on untill

@y-y = n<¡0!n>

This is similar to how factorial works. Reduce to n copies of the ¡n-1! separator all the way to ¡0!.

example: ¿1¡2!0?3

**Spoiler:**

Step 6: ¿0¡Y!0?n = <¿@¡Y--!@>n

From here on, (X¡Y!Z)-- means reduce applying rules 1 through 6

Chains: all separators are the same. examples: 3¡0!4¡0!5 and 100¡2!25¡2!50

Step 7: ¿P¡%!Q¡%!R?n = <¿P¡%!Q¡%!R--?>n

Step 8: ¿P¡%!Q¡%!0?n = <¿P¡%!Q--¡%!@?>n

Step 9: ¿P¡%!0¡%!0?n = <¿P--¡%!@¡%!@?>n

Step 10: ¿0¡%!0¡%!0?n = <¿@¡%!@?>n

Step 11: ¿0¡%!0¡%!0¡%!0?n = <¿@¡%!@¡%!@?>n // all-zero elements with all-equal separators eats a separator and an element

Below is chains where there are different separators, ¡L! denotes the lowest ranking one.

Step 12: ¿X¡%!P¡L!Q¡%!Y?n = <¿X¡%!(P¡L!Q)--¡%!Y?>n This ignores wether Y is 0 or not

Step 13: ¿X¡%!P¡L!Q¡L!R¡%!Y?n = <¿X¡%!(P¡L!Q¡L!R)--¡%!Y?>n This ignores wether Y is 0 or not

Step 14: ¿X¡%!P¡L!Q¡L!R¡%!Y¡L!Z¡%!W?n = <¿X¡%!P¡L!Q¡L!R¡%!(Y¡L!Z)--¡%!W?>n This ignores wether W is 0 or not

The "active" part or, the one to be reduced, is the lowest ranking and rightmost group.

Nesting. The same chain-handling and cascading rules apply to all kinds of separators.

Step 15: ¿0¡0¡0!0!0?n = <¿@¡@!@?>n Begin cascade at n<¡n!n>

Example: ¿0¡0¡0!0!0?3 = <¿@¡3!@¡3!@¡3!@>

Step 16: ¿X¡%!Y?n = <¿X¡%!Y--?>n

Step 17: ¿X¡%!0?n = <¿X--¡%!@?>n

Step 18: ¿0¡%!0?n = <¿@¡%--!@?>n

Step 15b: ¿0¡0¡0¡0!0!0!0?n = <¿@¡@¡@!@!@?>n

example: ¿0¡0¡0¡0!0!0!0?3= <¿@¡@¡3!@¡3!@¡3!@!@>n

All-zero elements (not separators) on the nested separators eat a level of nesting.

Step 19: ¿0¡0!¡0!0?n = <@¡@!@>n //

Step 20: ¿0¡#!¡%!0?n = <@¡#!¡(%)--!@>n % Can be nested

Step 21: ¿0¡%!¡0!0?n = <@¡(%)--!¡@!@>n % Can be nested

Step 19b: ¿0¡0!¡0!¡0!0?n = <@¡@!¡@!@>n

Edit: Hopefully made the wording clearer

Last edited by Daggoth on Sun Mar 08, 2015 8:08 pm UTC, edited 8 times in total.

### Re: Your number is, in fact, not bigger!

I'll give a notation for the SVO, but first: Vytron, do you prefer the notation to have numbers, or not have numbers? (like my box notation)

### Re: Your number is, in fact, not bigger!

Deedlit wrote:I know WarDaft could really dial things up as well if he wanted to.

That is indeed what the colon is for.

All Shadow priest spells that deal Fire damage now appear green.

Big freaky cereal boxes of death.

- Vytron
**Posts:**429**Joined:**Mon Oct 19, 2009 10:11 am UTC**Location:**The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P

### Re: Your number is, in fact, not bigger!

Okay, I'm having problems showing clear directions of how to find strings in my notation.

I'm supposed to have:

n,x,a

Where x is a string that remains the same when collapsing (i.e. n,x,a = n,x,a--)

To find where x ends, look at the highest element after the first , find the rightmost instance of this element, and find the first element that is as high as this element to the left of it, that's where x ends.

So:

n,0,0,0,{0},0,{0},{0},0,0

Find the highest element, that's a {0}.

Find the rightmost {0}, that's:

n,0,0,0,{0},0,{0},{0},0,0

Find the first {0} to the left of it, that's:

n,0,0,0,{0},0,{0},{0},0,0

So, that's where x ends, making:

x=,0,0,0,{0},0,{0} a=,{0},0,0

They're equivalent.

n,x,{0} means that x ends in a {0} (otherwise we'd have n,x,0,{0} or any number of 0s to the left of {0}), so having n,x,{0} means 0s don't exist to the left of {0} in a.

Suppose we have:

n,{0,0},{0}

In this case, x would be empty, because there's no element as high as {0,0} to the left of {0,0}, which makes:

x={0,0} a=,{0}

So by these rules:

n,{0,0},{0} = n,{0,0},<,0>

Yeah, but then, if we have:

n,x,{0},{0}

We ignore x and we have:

n,{0},{0}

And here, x={0} (the left one) and a={0} (the one on the right), so we return:

n,{0},<,0>

So we return:

n,x,{0},{0} = n,x,{0}<,0>

No, the progression is as follows, where each new line is bigger than the previous one (restricting to 3 elements):

{0,0,0}

{0,0,0},0

{0,0,0},0,0

{0,0,0},0,0,0 = {0,0,0},{0}

{0,0,0},{0},0

{0,0,0},{0},0,0

{0,0,0},{0},0,0,0 = {0,0,0},{0},{0}

{0,0,0},{0},{0},0

{0,0,0},{0},{0},0,0

{0,0,0},{0},{0},0,0,0 = {0,0,0},{0},{0},{0} = {0,0,0},{0,0}

{0,0,0},{0,0},0

{0,0,0},{0,0},0,0

{0,0,0},{0,0},0,0,0 = {0,0,0},{0,0},{0}

{0,0,0},{0,0},{0},0

{0,0,0},{0,0},{0},0,0

{0,0,0},{0,0},{0},0,0,0 = {0,0,0},{0,0},{0},{0}

{0,0,0},{0,0},{0},{0},0

{0,0,0},{0,0},{0},{0},0,0

{0,0,0},{0,0},{0},{0},0,0,0 = {0,0,0},{0,0},{0},{0},{0} = {0,0,0},{0,0},{0,0}

{0,0,0},{0,0},{0,0},{0,0} = {0,0,0},{0,0,0}

{0,0,0},{0,0,0},{0,0},{0,0},{0,0} = {0,0,0},{0,0,0},{0,0,0} = 0,{0,0,0}

That is:

n,0,{0,0,0} = n<,{0,0,0}>

You get a large number of {0,0,0}s, but now you don't ever have a lower element at the left of the {0,0,0}, so now:

n,{0,0,0},{0,0,0},{0,0,0}

{0,0,0} is the highest element so x ends at the penultimate {0,0,0}

n,x,{0,0,0} = n,x,<,0>,{0},{0,0}

So we get:

n,{0,0,0},{0,0,0},{0,0,0} = n,{0,0,0},{0,0,0},<,0>,{0},{0,0}

And this doesn't loop.

There's an alternative way to show my notation, where instead of adding {}s and 0s, there's equivalence with these values:

0=0

{0}=1

{0,0}=2

{0,0,0}=3

In this case mike's notation works for a+1.

There's a way to fix this though it does look ugly.

Instead of {}s, []s and ()s, we have these:

[x[ ]x]

Where:

[0[ = {

]0] = }

[0,0[ = [

]0,0] = ]

[0,0,0[ = (

]0,0,0] = )

And:

[0,0,0,0[ ]0,0,0,0]

Can be used as the separator for our purposes. And the entire notation can fit inside the x in [x[ ]x]. Unless you have a better idea.

@Daggoth: I think it looks better with numbers.

I'm supposed to have:

n,x,a

Where x is a string that remains the same when collapsing (i.e. n,x,a = n,x,a--)

To find where x ends, look at the highest element after the first , find the rightmost instance of this element, and find the first element that is as high as this element to the left of it, that's where x ends.

So:

n,0,0,0,{0},0,{0},{0},0,0

Find the highest element, that's a {0}.

Find the rightmost {0}, that's:

n,0,0,0,{0},0,{0},{0},0,0

Find the first {0} to the left of it, that's:

n,0,0,0,{0},0,{0},{0},0,0

So, that's where x ends, making:

x=,0,0,0,{0},0,{0} a=,{0},0,0

Deedlit wrote:Wait, that seems different from the rule

n,x,{0} = n,x<,0>

which is the rule that would seem to apply here.

They're equivalent.

n,x,{0} means that x ends in a {0} (otherwise we'd have n,x,0,{0} or any number of 0s to the left of {0}), so having n,x,{0} means 0s don't exist to the left of {0} in a.

Deedlit wrote:But isn't x either the rightmost element of n,0,x, or a string that contains the rightmost element? When is it the case that x is higher?

Suppose we have:

n,{0,0},{0}

In this case, x would be empty, because there's no element as high as {0,0} to the left of {0,0}, which makes:

x={0,0} a=,{0}

So by these rules:

n,{0,0},{0} = n,{0,0},<,0>

Deedlit wrote:Wait, the rules don't say anything about a eating part of b. It just says to take n,a,b--, so we should just reduce {0},{0} as if a did not exist.

Yeah, but then, if we have:

n,x,{0},{0}

We ignore x and we have:

n,{0},{0}

And here, x={0} (the left one) and a={0} (the one on the right), so we return:

n,{0},<,0>

So we return:

n,x,{0},{0} = n,x,{0}<,0>

Deedlit wrote:This is where it seems like it would go infinite.

Let's fix n at 3, and say we had 3,0,{0,0,0}. By the rule you just described above, this evaluates to

3,<<>,{0,0,0}> = 3<,{0,0,0}> = 3,{0,0,0}{0,0,0},{0,0,0}

Note that we now have more {0,0,0}'s than before. If we try to evaluate this again, we get something like

3<<,{0,0,0},{0,0,0}-->,{0,0,0}> = 3,{0,0,0},{0,0,0}--,{0,0,0},{0,0,0}--,{0,0,0},{0,0,0}--,{0,0,0},{0,0,0},{0,0,0}--,{0,0,0},{0,0,0}--,{0,0,0},{0,0,0}--,{0,0,0},{0,0,0},{0,0,0}--,{0,0,0},{0,0,0}--,{0,0,0},{0,0,0}--,{0,0,0}

where the {0,0,0}--'s are whatever {0,0,0} evaluates to. The point is that there are now more {0,0,0}'s than there were before, and there are going to keep being more and more.

No, the progression is as follows, where each new line is bigger than the previous one (restricting to 3 elements):

{0,0,0}

{0,0,0},0

{0,0,0},0,0

{0,0,0},0,0,0 = {0,0,0},{0}

{0,0,0},{0},0

{0,0,0},{0},0,0

{0,0,0},{0},0,0,0 = {0,0,0},{0},{0}

{0,0,0},{0},{0},0

{0,0,0},{0},{0},0,0

{0,0,0},{0},{0},0,0,0 = {0,0,0},{0},{0},{0} = {0,0,0},{0,0}

{0,0,0},{0,0},0

{0,0,0},{0,0},0,0

{0,0,0},{0,0},0,0,0 = {0,0,0},{0,0},{0}

{0,0,0},{0,0},{0},0

{0,0,0},{0,0},{0},0,0

{0,0,0},{0,0},{0},0,0,0 = {0,0,0},{0,0},{0},{0}

{0,0,0},{0,0},{0},{0},0

{0,0,0},{0,0},{0},{0},0,0

{0,0,0},{0,0},{0},{0},0,0,0 = {0,0,0},{0,0},{0},{0},{0} = {0,0,0},{0,0},{0,0}

{0,0,0},{0,0},{0,0},{0,0} = {0,0,0},{0,0,0}

{0,0,0},{0,0,0},{0,0},{0,0},{0,0} = {0,0,0},{0,0,0},{0,0,0} = 0,{0,0,0}

That is:

n,0,{0,0,0} = n<,{0,0,0}>

You get a large number of {0,0,0}s, but now you don't ever have a lower element at the left of the {0,0,0}, so now:

n,{0,0,0},{0,0,0},{0,0,0}

{0,0,0} is the highest element so x ends at the penultimate {0,0,0}

n,x,{0,0,0} = n,x,<,0>,{0},{0,0}

So we get:

n,{0,0,0},{0,0,0},{0,0,0} = n,{0,0,0},{0,0,0},<,0>,{0},{0,0}

And this doesn't loop.

[X,b,Y,a]-- = [X,b,<Y--, a>] when b>=a, all terms in Y are < a.

[X,b,a+1]-- = [X,b, a--, a] when b> a

what is a+1? So far everything is composed of 0's.

There's an alternative way to show my notation, where instead of adding {}s and 0s, there's equivalence with these values:

0=0

{0}=1

{0,0}=2

{0,0,0}=3

In this case mike's notation works for a+1.

Unfortunately spaces are nondirectional, so it is not clear if a space means you are enclosing the part to the right or unenclosing the part to the left.

There's a way to fix this though it does look ugly.

Instead of {}s, []s and ()s, we have these:

[x[ ]x]

Where:

[0[ = {

]0] = }

[0,0[ = [

]0,0] = ]

[0,0,0[ = (

]0,0,0] = )

And:

[0,0,0,0[ ]0,0,0,0]

Can be used as the separator for our purposes. And the entire notation can fit inside the x in [x[ ]x]. Unless you have a better idea.

@Daggoth: I think it looks better with numbers.

### Re: Your number is, in fact, not bigger!

Vytron wrote:Okay, I'm having problems showing clear directions of how to find strings in my notation.

I'm supposed to have:

n,x,a

Where x is a string that remains the same when collapsing (i.e. n,x,a = n,x,a--)

To find where x ends, look at the highest element after the first , find the rightmost instance of this element, and find the first element that is as high as this element to the left of it, that's where x ends.

So:

n,0,0,0,{0},0,{0},{0},0,0

Find the highest element, that's a {0}.

Find the rightmost {0}, that's:

n,0,0,0,{0},0,{0},{0},0,0

Find the first {0} to the left of it, that's:

n,0,0,0,{0},0,{0},{0},0,0

So, that's where x ends, making:

x=,0,0,0,{0},0,{0} a=,{0},0,0

That's quite clear, you just needed to state it.

They're equivalent.

n,x,{0} means that x ends in a {0} (otherwise we'd have n,x,0,{0} or any number of 0s to the left of {0}), so having n,x,{0} means 0s don't exist to the left of {0} in a.

So if I read you right, the rule

n,0,x = n,x<,x>

could be expanded to

n,y,0,x = n,y,x<,x>

is that correct?

[X,b,Y,a]-- = [X,b,<Y--, a>] when b>=a, all terms in Y are < a.

[X,b,a+1]-- = [X,b, a--, a] when b> a

If these rules apply generally, then I will try to analyze your notation a bit

0 = 1

0,0 = 2

0,0,0 = 3

{0} = ω

{0},0 = ω+1

{0},0,0 = ω+2

{0},0,0,0 = ω+3

{0},{0} = ω2

{0},{0},{0} = ω3

{0},{0},{0},{0} = ω4

0,{0} = ω^2

0,{0},0,{0} = ω^2 2

0,{0},0,(0),0,{0} = ω^2 3

0,0,{0} = ω^3

0,0,0,{0} = ω^4

0,0,0,0,{0} = ω^5

{0,0} = ω^ω

{0,0},{0,0} = ω^ω 2

0,{0,0} = ω^(ω+1)

0,0,{0,0} = ω^(ω+2)

0,0,0,{0,0} = ω^(ω+3)

{0},{0,0} = ω^(ω2)

{0},0,{0,0} = ω^(ω2+1)

{0},0,0,{0,0} = ω^(ω2+2)

{0},{0},{0,0} = ω^(ω3)

{0],{0},{0},{0,0} = ω^(ω4)

0,{0},{0,0} = ω^(ω^2)

0,0,{0},{0,0} = ω^(ω^3)

0,0,0,{0},{0,0} = ω^(ω^4)

{0,0,0} = ω^(ω^ω)

{0,0,0,0} = ω^ω^ω^ω

{0,0,0,0,0} = ω^ω^ω^ω^ω

{{0}} = ε_0

{{0}},{{0}} = ε_0 2

0,{{0}} = ε_0 ω

0,0,{0}} = ε_0 ω^2

{0},{{0}} = ε_0 ω^ω

{0,0}, {{0}} = ε_0 ω^ω^ω

{0,0,0}, {{0}} = ε_0 ω^ω^ω^ω

at this point there doesn't seem to be anything that evaluates to {<0,>},{{0}}, but I assume the next notation would be {{0},0}.

How do you evaluate {X} where X is something higher than {0}?

### Re: Your number is, in fact, not bigger!

Okay, here's a notation that's at the level of the SVO. As Vytron prefers numbers, I'll make it have numbers, even though it makes it a little more complicated.

Notations will consist of a number followed by an expression. An expression will be a string consisting of numbers, commas, and parentheses. Parentheses will of course be matching. Commas can only be inside parentheses. Numbers never immediately follow a right parenthesis. Also, having commas right before a right parentheses does nothing, so we will remove them whenever that happens.

To reduce an expression. go to the leftmost number.

If it is the leftmost element of the expression, reduce it by 1, or remove it if it is 0.

Otherwise find the pair of parentheses immediately enclosing the leftmost number, and call it the active pair.

If the active pair does not contain a comma:

-reduce the leftmost number by 1 if it is greater than 0

-if 0 is not the only thing in the active pair, remove it

-otherwise, replace (0) with n

If the active pair does contain a comma:

-if there are commas to the left of the leftmost number, we use the rule

(C , nX)-- = <(C><,n-- X)> where C is a string of commas, and X can be anything (remember n-- is n-1 if n > 0, or nothing if n = 0)

-if there are no commas to the left of the leftmost number, then we use the rule

(nX C , Y)-- = <(n X C> <(n-- X C, Y)> <, Y--)> {where C is a string of commas, and X and Y can be anything}

Or, we can just use a set of rules:

n| = n

n|X = n+1|X--

0X -- = X

{n+1}X -- = nX

(X)Y -- = <(X--)> Y

(0)X -- = nX

(C , X)-- = <(C><, X--)>

(X C, Y)-- = <(X C> <(X-- C, Y)> <, Y--)>

Note: the two rules sets are not exactly the same, I adjusted them so that they would be simpler to describe.

Examples: (using second rule set)

3 | 6 (4,(3),(2,4))(1,5) = 4 | 5 (4,(3),(2,4))(1,5) = 5 | 4 (4,(3),(2,4))(1,5) = 6 | 3 (4,(3),(2,4))(1,5) = ...

4 | (3, ((2,5)(4))) = 5 | (2, ((2,5)(4))) (2, ((2,5)(4))) (2, ((2,5)(4))) (2, ((2,5)(4)))

3 | (((0(5)(2)))) = 4 | <(<(<(0(5)(2))>)>)> = 4 | ((((5)(2))((5)(2))((5)(2)))(((5)(2))((5)(2))((5)(2)))(((5)(2))((5)(2))((5)(2)))((((5)(2))((5)(2))((5)(2)))(((5)(2))((5)(2))((5)(2)))(((5)(2))((5)(2))((5)(2)))((((5)(2))((5)(2))((5)(2)))(((5)(2))((5)(2))((5)(2)))(((5)(2))((5)(2))((5)(2)))

3 | ((1, , , (1))) = 4 | <(<(1 , ,> <(0, , (1))> <, (0)(0)(0)))> = ( (1 , ,(1 , ,(1 , ,(0, , , (1))(0, , , (1))(0, , , (1)), (0)(0)(0)), (0)(0)(0)), (0)(0)(0)) )( (1 , ,(1 , ,(1 , ,(0, , , (1))(0, , , (1))(0, , , (1)), (0)(0)(0)), (0)(0)(0)), (0)(0)(0)) )( (1 , ,(1 , ,(1 , ,(0, , , (1))(0, , , (1))(0, , , (1)), (0)(0)(0)), (0)(0)(0)), (0)(0)(0)) )

3 | ((, , , (1(4)))) = 4 | <( <(, ,> <, (0(4))(0(4))(0(4)))> )> = 4 | (, ,(, ,(, , , (0(4))(0(4))(0(4))), (0(4))(0(4))(0(4))), (0(4))(0(4))(0(4))) ) (, ,(, ,(, , , (0(4))(0(4))(0(4))), (0(4))(0(4))(0(4))), (0(4))(0(4))(0(4))) ) (, ,(, ,(, , , (0(4))(0(4))(0(4))), (0(4))(0(4))(0(4))), (0(4))(0(4))(0(4))) )

If there is any confusion, please ask for clarification.

Notations will consist of a number followed by an expression. An expression will be a string consisting of numbers, commas, and parentheses. Parentheses will of course be matching. Commas can only be inside parentheses. Numbers never immediately follow a right parenthesis. Also, having commas right before a right parentheses does nothing, so we will remove them whenever that happens.

To reduce an expression. go to the leftmost number.

If it is the leftmost element of the expression, reduce it by 1, or remove it if it is 0.

Otherwise find the pair of parentheses immediately enclosing the leftmost number, and call it the active pair.

If the active pair does not contain a comma:

-reduce the leftmost number by 1 if it is greater than 0

-if 0 is not the only thing in the active pair, remove it

-otherwise, replace (0) with n

If the active pair does contain a comma:

-if there are commas to the left of the leftmost number, we use the rule

(C , nX)-- = <(C><,n-- X)> where C is a string of commas, and X can be anything (remember n-- is n-1 if n > 0, or nothing if n = 0)

-if there are no commas to the left of the leftmost number, then we use the rule

(nX C , Y)-- = <(n X C> <(n-- X C, Y)> <, Y--)> {where C is a string of commas, and X and Y can be anything}

Or, we can just use a set of rules:

n| = n

n|X = n+1|X--

0X -- = X

{n+1}X -- = nX

(X)Y -- = <(X--)> Y

(0)X -- = nX

(C , X)-- = <(C><, X--)>

(X C, Y)-- = <(X C> <(X-- C, Y)> <, Y--)>

Note: the two rules sets are not exactly the same, I adjusted them so that they would be simpler to describe.

Examples: (using second rule set)

3 | 6 (4,(3),(2,4))(1,5) = 4 | 5 (4,(3),(2,4))(1,5) = 5 | 4 (4,(3),(2,4))(1,5) = 6 | 3 (4,(3),(2,4))(1,5) = ...

4 | (3, ((2,5)(4))) = 5 | (2, ((2,5)(4))) (2, ((2,5)(4))) (2, ((2,5)(4))) (2, ((2,5)(4)))

3 | (((0(5)(2)))) = 4 | <(<(<(0(5)(2))>)>)> = 4 | ((((5)(2))((5)(2))((5)(2)))(((5)(2))((5)(2))((5)(2)))(((5)(2))((5)(2))((5)(2)))((((5)(2))((5)(2))((5)(2)))(((5)(2))((5)(2))((5)(2)))(((5)(2))((5)(2))((5)(2)))((((5)(2))((5)(2))((5)(2)))(((5)(2))((5)(2))((5)(2)))(((5)(2))((5)(2))((5)(2)))

3 | ((1, , , (1))) = 4 | <(<(1 , ,> <(0, , (1))> <, (0)(0)(0)))> = ( (1 , ,(1 , ,(1 , ,(0, , , (1))(0, , , (1))(0, , , (1)), (0)(0)(0)), (0)(0)(0)), (0)(0)(0)) )( (1 , ,(1 , ,(1 , ,(0, , , (1))(0, , , (1))(0, , , (1)), (0)(0)(0)), (0)(0)(0)), (0)(0)(0)) )( (1 , ,(1 , ,(1 , ,(0, , , (1))(0, , , (1))(0, , , (1)), (0)(0)(0)), (0)(0)(0)), (0)(0)(0)) )

3 | ((, , , (1(4)))) = 4 | <( <(, ,> <, (0(4))(0(4))(0(4)))> )> = 4 | (, ,(, ,(, , , (0(4))(0(4))(0(4))), (0(4))(0(4))(0(4))), (0(4))(0(4))(0(4))) ) (, ,(, ,(, , , (0(4))(0(4))(0(4))), (0(4))(0(4))(0(4))), (0(4))(0(4))(0(4))) ) (, ,(, ,(, , , (0(4))(0(4))(0(4))), (0(4))(0(4))(0(4))), (0(4))(0(4))(0(4))) )

If there is any confusion, please ask for clarification.

- Vytron
**Posts:**429**Joined:**Mon Oct 19, 2009 10:11 am UTC**Location:**The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P

### Re: Your number is, in fact, not bigger!

Thanks Deedlit. Yes, mike-l has done extensive analysis of my notation and shown what ordinals are reached where. Here's a link to his post about it on the epsilon numbers reached, and a link on how it reaches z_0. Though, my notation looked different back then - ,s were +s, and I used integers instead of nested 0s, but the same concepts should apply. Here's a latter summary, in there what was 00 is now [0-0], 000 is now [0-0,0], and so on.

To know what follows after {{0}} one just keeps going, so we have:

n,{{0},0} = n,{0<,0>},{{0}}

n,0,{{0},0} = n<,{{0},0}>

n,{0},{{0},0} = n<<,0>{{0},0}>

n,{0,0},{{0},0} = n<<<,0>,{0},>{{0},0}>

n,{0,0,0},{{0},0} = n<<<<,0>,{0}>,{0,0},>{{0},0}>

n,{{0}},{{0},0} = n<<{<,0>},>{{0},0}>

n,{{0},0,0} = n<<,{0<,0>},{{0}},>{{0},0}>

n,{{0},0,0,0} = n<<<,{0<,0>},{{0}},>{{0},0}>{{0},0,0}>

n,{{0},0,0,0,0} = n<<<<,{0<,0>},{{0}},>{{0},0}>{{0},0,0}>{{0},0,0,0}>

n,{{0},{0}} = n,{{0}<,0>}

And at this point the notation in {x} follows the original progression (i.e. n,{0,{0}} = n,{{0}<,{0}>}), and:

n,[0-0] = n,<{>0<}> should reach ζ_0.

--------

And now, I've thought of a way to solve the problem of "running out of directional enclosers". For this, I'll need to rewrite a bit of my notation. People have complained that I rewrite my notation every 3 days making all the effort to understand it null, but I have claimed all my notations have been equivalent and interchangeable, so analysis for one applies for all of them.

Use the rules:

n,X,b,Y,a -- = n,X,b,<Y--, a> when b>=a, all terms in Y are < a.

n,X,b,a+1 -- = X,b, a--, a when b> a

Where a+1 = a means ,0 = nothing (i.e. this drops a 0)

Now, instead of introducing []s, make:

n,{0-0} = n,<{>0<}>

n,{0-x} = Follows the rules of above

n,{0,0-0} = n,<{0->{0}<}>

n,{0,0,0-0} = n,<{0,0->{0}<}>

n,{x-0} = Follows the rules of above

n,{0-0-0}= n,<{>{0}<-0}>

n,{0-0,0-0} = n,<{0-0>-{0}<}>

n,{0-x-0} = Follows the rules of above

n,{0,0-x-0} = n,<{0->{0}<-0}>

n,{x-0-0} = Follows the rules of above

n,{0-0-0-0}= n,<{>{0}<-0-0}>

n,{0-0-0-0-0}= n,<{>{0}<-0-0-0}>

n,{0-0-0-0-0-0}= n,<{>{0}<-0-0-0-0}>

Basically, I had no reason whatsoever to switch to square boxes when I did. I did it because, old notations that used integers>0 didn't use {}s, so when I introduced bunch of 0s instead of integers I kept them, but {}s work because if there's a - in {X} you know it's a larger element than one without, and more -s always are a higher element.

Now, instead of going for integers higher than 0, I introduce the square brackets:

n,[0] = n,{0<-0>}

n,[0,0] = <(>n<),[0]>

n,[0,0,0] = <(>n<),[0,0]>

n,[{0}] = n,[0<,0>]

Basically, now I can fit my entire notation in n,[X], up until:

n,[[0]] = n,[{0<-0>}]

n,[[0],0] = <(>n<),[[0]]>

n,[[0],0,0] = <(>n<),[[0],0]>

n,[[0],{0}] = n,[[0]<,0>]

Entire notation in n,[X]

There should be this correspondence:

n,[0] = φ(1@ω)

n,[[0]] = φ(1@ω+1)

n,[[0],[0]] = φ(1@ω+2)

n,[0,[0]] = n[[0]<,[0]>] = φ(1@ω2)

n,[0,[0],0,[0]] = n[0,[0]<,[0]>] = φ(1@ω3)

And now I send:

3,[0,[0],0,[0]]

Ninja'd

Thanks for posting your notation Deedlit - I think it would really help if you posted equivalences with the FGH to see how it grows.

How do you evaluate {X} where X is something higher than {0}?

To know what follows after {{0}} one just keeps going, so we have:

n,{{0},0} = n,{0<,0>},{{0}}

n,0,{{0},0} = n<,{{0},0}>

n,{0},{{0},0} = n<<,0>{{0},0}>

n,{0,0},{{0},0} = n<<<,0>,{0},>{{0},0}>

n,{0,0,0},{{0},0} = n<<<<,0>,{0}>,{0,0},>{{0},0}>

n,{{0}},{{0},0} = n<<{<,0>},>{{0},0}>

n,{{0},0,0} = n<<,{0<,0>},{{0}},>{{0},0}>

n,{{0},0,0,0} = n<<<,{0<,0>},{{0}},>{{0},0}>{{0},0,0}>

n,{{0},0,0,0,0} = n<<<<,{0<,0>},{{0}},>{{0},0}>{{0},0,0}>{{0},0,0,0}>

n,{{0},{0}} = n,{{0}<,0>}

And at this point the notation in {x} follows the original progression (i.e. n,{0,{0}} = n,{{0}<,{0}>}), and:

n,[0-0] = n,<{>0<}> should reach ζ_0.

--------

And now, I've thought of a way to solve the problem of "running out of directional enclosers". For this, I'll need to rewrite a bit of my notation. People have complained that I rewrite my notation every 3 days making all the effort to understand it null, but I have claimed all my notations have been equivalent and interchangeable, so analysis for one applies for all of them.

Use the rules:

n,X,b,Y,a -- = n,X,b,<Y--, a> when b>=a, all terms in Y are < a.

n,X,b,a+1 -- = X,b, a--, a when b> a

Where a+1 = a means ,0 = nothing (i.e. this drops a 0)

Now, instead of introducing []s, make:

n,{0-0} = n,<{>0<}>

n,{0-x} = Follows the rules of above

n,{0,0-0} = n,<{0->{0}<}>

n,{0,0,0-0} = n,<{0,0->{0}<}>

n,{x-0} = Follows the rules of above

n,{0-0-0}= n,<{>{0}<-0}>

n,{0-0,0-0} = n,<{0-0>-{0}<}>

n,{0-x-0} = Follows the rules of above

n,{0,0-x-0} = n,<{0->{0}<-0}>

n,{x-0-0} = Follows the rules of above

n,{0-0-0-0}= n,<{>{0}<-0-0}>

n,{0-0-0-0-0}= n,<{>{0}<-0-0-0}>

n,{0-0-0-0-0-0}= n,<{>{0}<-0-0-0-0}>

Basically, I had no reason whatsoever to switch to square boxes when I did. I did it because, old notations that used integers>0 didn't use {}s, so when I introduced bunch of 0s instead of integers I kept them, but {}s work because if there's a - in {X} you know it's a larger element than one without, and more -s always are a higher element.

Now, instead of going for integers higher than 0, I introduce the square brackets:

n,[0] = n,{0<-0>}

n,[0,0] = <(>n<),[0]>

n,[0,0,0] = <(>n<),[0,0]>

n,[{0}] = n,[0<,0>]

Basically, now I can fit my entire notation in n,[X], up until:

n,[[0]] = n,[{0<-0>}]

n,[[0],0] = <(>n<),[[0]]>

n,[[0],0,0] = <(>n<),[[0],0]>

n,[[0],{0}] = n,[[0]<,0>]

Entire notation in n,[X]

There should be this correspondence:

n,[0] = φ(1@ω)

n,[[0]] = φ(1@ω+1)

n,[[0],[0]] = φ(1@ω+2)

n,[0,[0]] = n[[0]<,[0]>] = φ(1@ω2)

n,[0,[0],0,[0]] = n[0,[0]<,[0]>] = φ(1@ω3)

And now I send:

3,[0,[0],0,[0]]

Ninja'd

Thanks for posting your notation Deedlit - I think it would really help if you posted equivalences with the FGH to see how it grows.

### Re: Your number is, in fact, not bigger!

Okay, here are ordinal equivalences for my SVO notation:

n = n

(0) = ω

n(0) = ω+n

(0)(0) = ω2

(0)(0)(0) = ω3

(1) = ω^2

(2) = ω^3

(3) = ω^4

((0)) = ω^ω

(0(0)) = ω^(ω+1)

((0)(0)) = ω^(ω2)

((0)(0)(0)) = ω^(ω3)

((1)) = ω^(ω^2)

((2)) = ω^(ω^3)

((3)) = ω^(ω^4)

(((0))) = ω^(ω^ω)

((((0)))) = ω^(ω^(ω^ω))

(,0) = e_0

0(,0) = e_0 + 1

(0)(,0) = e_0 + ω

(0,0)(,0) = e_0 2

(0(,0)) = e_0 ω

(1(,0)) = e_0 ω^2

((0)(,0)) = e_0 ω^ω

((,0)(,0)) = e_0^2

((,0)(,0)(,0)) = e_0^3

((0(,0))) = e_0^ω = ω^ω^(e_0+1)

((1(,0))) = ω^ω^(e_0 + 2)

(((0)(,0))) = ω^ω^(e_0 + ω)

(((,0)(,0))) = ω^ω^(e_0 + e_0)

(((0(,0)))) = ω^ω^ω^(e_0+1)

(0,0) = e_1

(1,0) = e_2

(2,0) = e_3

((0),0) = e_ω

((,0),0) = e_e_0

(((,0),0),0) = e_e_e_0

(,1) = z_0

(,1)(,1) = z_0 2

(0(,1)) = ω^(z_0 + 1)

((0)(,1)) = ω^(z_0 + ω)

((,0)(,1)) = ω^(z_0 + e_0)

((,1)(,1)) = ω^(z_0 2)

(((,1)(,1))) = ω^ω^(z_0 2)

((,1),0) = e_(z_0+1)

(0(,1),0) = e_(z_0 + 2)

((,1)(,1),0) = e_(z_0 2)

(((,1),0),0) = e_e_(z_0 + 1)

(0,1) = z_1

(1,1) = z_2

(2,1) = z_3

(,2) = phi(3,0)

(0,2) = phi(3,1)

(1,2) = phi(3,2)

(,3) = phi(4,0)

(,4) = phi(5,0)

(,(0)) = phi(ω,0)

(,0(0)) = phi(ω+1,0)

(,(0)(0)) = phi(ω2,0)

(,(,0)) = phi(e_0,0)

(,(0,0)) = phi(e_1,0)

(,(,1)) = phi(z_0,0)

(,(,2)) = phi(phi(3,0),0)

(,(,(0))) = phi(phi(ω,0),0)

(,(,(,0))) = phi(phi(e_0,0),0)

(,(,(,(,0)))) = phi(phi(phi(e_0,0),0),0)

(,,0) = phi(1,0,0)

(,,1) = phi(2,0,0)

(,,(0)) = phi(ω,0,0)

(,,(,0)) = phi(e_0,0,0)

(,,(,,0)) = phi(phi(1,0,0),0,0)

(,,,0) = phi(1,0,0,0)

(,,,,0) = phi(1,0,0,0,0)

Limit is SVO.

n = n

(0) = ω

n(0) = ω+n

(0)(0) = ω2

(0)(0)(0) = ω3

(1) = ω^2

(2) = ω^3

(3) = ω^4

((0)) = ω^ω

(0(0)) = ω^(ω+1)

((0)(0)) = ω^(ω2)

((0)(0)(0)) = ω^(ω3)

((1)) = ω^(ω^2)

((2)) = ω^(ω^3)

((3)) = ω^(ω^4)

(((0))) = ω^(ω^ω)

((((0)))) = ω^(ω^(ω^ω))

(,0) = e_0

0(,0) = e_0 + 1

(0)(,0) = e_0 + ω

(0,0)(,0) = e_0 2

(0(,0)) = e_0 ω

(1(,0)) = e_0 ω^2

((0)(,0)) = e_0 ω^ω

((,0)(,0)) = e_0^2

((,0)(,0)(,0)) = e_0^3

((0(,0))) = e_0^ω = ω^ω^(e_0+1)

((1(,0))) = ω^ω^(e_0 + 2)

(((0)(,0))) = ω^ω^(e_0 + ω)

(((,0)(,0))) = ω^ω^(e_0 + e_0)

(((0(,0)))) = ω^ω^ω^(e_0+1)

(0,0) = e_1

(1,0) = e_2

(2,0) = e_3

((0),0) = e_ω

((,0),0) = e_e_0

(((,0),0),0) = e_e_e_0

(,1) = z_0

(,1)(,1) = z_0 2

(0(,1)) = ω^(z_0 + 1)

((0)(,1)) = ω^(z_0 + ω)

((,0)(,1)) = ω^(z_0 + e_0)

((,1)(,1)) = ω^(z_0 2)

(((,1)(,1))) = ω^ω^(z_0 2)

((,1),0) = e_(z_0+1)

(0(,1),0) = e_(z_0 + 2)

((,1)(,1),0) = e_(z_0 2)

(((,1),0),0) = e_e_(z_0 + 1)

(0,1) = z_1

(1,1) = z_2

(2,1) = z_3

(,2) = phi(3,0)

(0,2) = phi(3,1)

(1,2) = phi(3,2)

(,3) = phi(4,0)

(,4) = phi(5,0)

(,(0)) = phi(ω,0)

(,0(0)) = phi(ω+1,0)

(,(0)(0)) = phi(ω2,0)

(,(,0)) = phi(e_0,0)

(,(0,0)) = phi(e_1,0)

(,(,1)) = phi(z_0,0)

(,(,2)) = phi(phi(3,0),0)

(,(,(0))) = phi(phi(ω,0),0)

(,(,(,0))) = phi(phi(e_0,0),0)

(,(,(,(,0)))) = phi(phi(phi(e_0,0),0),0)

(,,0) = phi(1,0,0)

(,,1) = phi(2,0,0)

(,,(0)) = phi(ω,0,0)

(,,(,0)) = phi(e_0,0,0)

(,,(,,0)) = phi(phi(1,0,0),0,0)

(,,,0) = phi(1,0,0,0)

(,,,,0) = phi(1,0,0,0,0)

Limit is SVO.

- Vytron
**Posts:**429**Joined:**Mon Oct 19, 2009 10:11 am UTC**Location:**The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P

### Re: Your number is, in fact, not bigger!

Thanks, that makes it very clear.

### Re: Your number is, in fact, not bigger!

I was wondering whether there could be an issue with the distinction between, say,

0-0 , 0-0

and

0 - 0.0 - 0

but I guess one can figure it out by looking at the structure of the {}'s. I'm not totally sure though.

Okay, so continuing the analysis of Vytron's notation:

{{0}, 0} = e_0^2

{{0}, 0} {{0}, 0} = e_0^2 2

0, {{0}, 0} = e_0^2 ω

{0}, {{0}, 0} = e_0^2 ω^ω

{{0}}, {{0}, 0} = e_0^3

{{0}}, 0, {{0},0} = e_0^3 ω

{{0}}, {{0}}, {{0}, 0} = e_0^4

{{0}}, {{0}}, {{0}}, {{0}, 0} = e_0^5

0,{{0}},{{0},0} = e_0^ω = ω^ω^(e_0+1)

0,{{0}},0,{{0},0} = ω^(ω^(e_0+1) + 1)

0,{{0}},{{0}},{{0},0} = ω^(ω^(e_0+1) + e_0)

0,{{0}},0,{{0}},{{0},0} = ω^(ω^(e_0+1) 2)

0,0,{{0}},{{0},0} = ω^(ω^(e_0 + 2))

0,0,0,{{0}},{{0},0} = ω^(ω^(e_0 + 3))

{0},{{0}},{{0},0} = ω^(ω^(e_0 + ω))

{{0},0,0} = ω^ω^(e_0 2)

{{0},0,0,0} = ω^ω^ω^(e_0 2)

{{0},{0}} = e_1

{{0},{0},{0}} = e_2

{{0},{0},{0},{0}} = e_3

{0, {0}} = e_ω

{0, {0}, 0} = e_ω^2

{0, {0}, {0}} = e_{ω+1}

{0, {0}, 0, {0}} = e_{ω2}

{0, 0, {0}} = e_{ω^2}

{0, 0, 0, {0}} = e_{ω^3}

{{0,0}} = e_{ω^ω}

{{0,0,0}} = e_{ω^(ω^ω)}

{{{0}}} = e_e_0

{{{{0}}}} = e_e_e_0

{0-0} = z_0

{0-0,0} = z_1

{0-0,0,0} = z_2

{0-{0}} = z_ω

{0-{{0}} = z_e_0

{0-{0-0}} = z_z_0

{0-{0-{0-0}}} = z_z_z_0

{1-0} = phi(3,0)

{1-1} = phi(3,1)

{2-0} = phi(4,0)

{{0}-0} = phi(ω,0)

{{0-0}-0} = phi(z_0, 0)

{0-0-0} = phi(1,0,0)

{0-0-0-0} = phi (1,0,0,0)

So yeah, I believe the limit of the - notation can be the SVO - IF the notation is defined correctly. I say this because because the definition Vytron provided is not quite complete, because the instructions "use the previous rules" doesn't make it quite clear how to handle them. To help me understand better, could you answer what

3,{{0-0}-{0,0}-0,0-0-0,0,0} evaluates to?

Or

3, {{0-0}-{0,0},0-{0,0}-0-0-0,0,0} ?

That is actually quite weak - remember iteration is just one level of the fast growing hierarchy, so this would just be

[0,0] = SVO+1

[0,0,0] = SVO+2

which I'm sure is not what you want.

0-0 , 0-0

and

0 - 0.0 - 0

but I guess one can figure it out by looking at the structure of the {}'s. I'm not totally sure though.

Okay, so continuing the analysis of Vytron's notation:

{{0}, 0} = e_0^2

{{0}, 0} {{0}, 0} = e_0^2 2

0, {{0}, 0} = e_0^2 ω

{0}, {{0}, 0} = e_0^2 ω^ω

{{0}}, {{0}, 0} = e_0^3

{{0}}, 0, {{0},0} = e_0^3 ω

{{0}}, {{0}}, {{0}, 0} = e_0^4

{{0}}, {{0}}, {{0}}, {{0}, 0} = e_0^5

0,{{0}},{{0},0} = e_0^ω = ω^ω^(e_0+1)

0,{{0}},0,{{0},0} = ω^(ω^(e_0+1) + 1)

0,{{0}},{{0}},{{0},0} = ω^(ω^(e_0+1) + e_0)

0,{{0}},0,{{0}},{{0},0} = ω^(ω^(e_0+1) 2)

0,0,{{0}},{{0},0} = ω^(ω^(e_0 + 2))

0,0,0,{{0}},{{0},0} = ω^(ω^(e_0 + 3))

{0},{{0}},{{0},0} = ω^(ω^(e_0 + ω))

{{0},0,0} = ω^ω^(e_0 2)

{{0},0,0,0} = ω^ω^ω^(e_0 2)

{{0},{0}} = e_1

{{0},{0},{0}} = e_2

{{0},{0},{0},{0}} = e_3

{0, {0}} = e_ω

{0, {0}, 0} = e_ω^2

{0, {0}, {0}} = e_{ω+1}

{0, {0}, 0, {0}} = e_{ω2}

{0, 0, {0}} = e_{ω^2}

{0, 0, 0, {0}} = e_{ω^3}

{{0,0}} = e_{ω^ω}

{{0,0,0}} = e_{ω^(ω^ω)}

{{{0}}} = e_e_0

{{{{0}}}} = e_e_e_0

{0-0} = z_0

{0-0,0} = z_1

{0-0,0,0} = z_2

{0-{0}} = z_ω

{0-{{0}} = z_e_0

{0-{0-0}} = z_z_0

{0-{0-{0-0}}} = z_z_z_0

{1-0} = phi(3,0)

{1-1} = phi(3,1)

{2-0} = phi(4,0)

{{0}-0} = phi(ω,0)

{{0-0}-0} = phi(z_0, 0)

{0-0-0} = phi(1,0,0)

{0-0-0-0} = phi (1,0,0,0)

So yeah, I believe the limit of the - notation can be the SVO - IF the notation is defined correctly. I say this because because the definition Vytron provided is not quite complete, because the instructions "use the previous rules" doesn't make it quite clear how to handle them. To help me understand better, could you answer what

3,{{0-0}-{0,0}-0,0-0-0,0,0} evaluates to?

Or

3, {{0-0}-{0,0},0-{0,0}-0-0-0,0,0} ?

n,[0] = n,{0<-0>}

n,[0,0] = <(>n<),[0]>

n,[0,0,0] = <(>n<),[0,0]>

n,[{0}] = n,[0<,0>]

That is actually quite weak - remember iteration is just one level of the fast growing hierarchy, so this would just be

[0,0] = SVO+1

[0,0,0] = SVO+2

which I'm sure is not what you want.

- Vytron
**Posts:**429**Joined:**Mon Oct 19, 2009 10:11 am UTC**Location:**The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P

### Re: Your number is, in fact, not bigger!

Deedlit wrote:3,{{0-0}-{0,0}-0,0-0-0,0,0} evaluates to?

If the leftmost element is not 0 then you can basically ignore everything before the rightmost -, so we have:

3,{x-0,0,0} = 3,{{{{{x-0,0}}}}} = 3,{{{{{{{{{x-0}}}}},{x-0,0}}}}

Now, the rightmost element of {x-0} is a 0, so we look at the element before that:

{x-0-0}

It's also a 0, so we look at the element before that.

{x-0,0-0-0}

So we reduce this element and return:

{x-0,0-0-0} = {{{{{x-0-{x-0-{x-0-{x-0-{x-0-0-0}-0}-0}-0}-0}}}}}

And make n+1 copies of the string

3,{{{{{{{{{x-0}}}}},{x-0,0}}}} = 3,{{{{{{{{{{{{{x-0-{x-0-{x-0-{x-0-{x-0-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0}}}}

To reduce {x-0-0-0} we look at the next element:

{x-{0,0}-0-0-0}

{0,0} reduces to 0,0,0,0,{0}, so:

{x-{0,0}-0-0-0} = {{{{{{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0-0}-0-0}-0-0}-0-0}-0-0}}}}}}

And make n+1 copies of the string:

3,{{{{{{{{{x-0-{x-0-{x-0-{x-0-{x-0-0-0}-0}-0}-0}-0}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0}}}} = 3,{{{{{{{{{x-0-{x-0-{x-0-{x-0-{x-0-0-0}-0}-0}-0}-0}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0-0}-0-0}-0-0}-0-0}-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0-0}-0-0}-0-0}-0-0}-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0-0}-0-0}-0-0}-0-0}-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0-0}-0-0}-0-0}-0-0}-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0-0}-0-0}-0-0}-0-0}-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0}}}}

{x-0,0,0,0,{0}-0-0-0} reduces to:

{{{{{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-0-0-0}-0-0}-0-0}-0-0}-0-0}}}}}

And make n+1 copies of the string:

3,{{{{{{{{{x-0-{x-0-{x-0-{x-0-{x-0-0-0}-0}-0}-0}-0}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0-0}-0-0}-0-0}-0-0}-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0-0}-0-0}-0-0}-0-0}-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0-0}-0-0}-0-0}-0-0}-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0-0}-0-0}-0-0}-0-0}-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0-0}-0-0}-0-0}-0-0}-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0}}}} = 3,{{{{{{{{{x-0-{x-0-{x-0-{x-0-{x-0-0-0}-0}-0}-0}-0}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0-0}-0-0}-0-0}-0-0}-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0-0}-0-0}-0-0}-0-0}-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0-0}-0-0}-0-0}-0-0}-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0-0}-0-0}-0-0}-0-0}-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-0-0-0}-0-0}-0-0}-0-0}-0-0}-0-0}-0-0}-0-0}-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-0-0-0}-0-0}-0-0}-0-0}-0-0}-0-0}-0-0}-0-0}-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-0-0-0}-0-0}-0-0}-0-0}-0-0}-0-0}-0-0}-0-0}-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0},{{{{{x-0-{x-0-{x-0-{x-0-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-0-0-0}-0-0}-0-0}-0-0}-0-0}-0-0}-0-0}-0-0}-0-0}-0}-0}-0}-0}}}}}}}}},{x-0,0}}}}

Reduce:

{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-0-0-0} = {{{{{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,{0},0,0,{0},0,0,{0},0,0,{0},0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,{0},0,0,{0},0,0,{0},0,0,{0},0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,{0},0,0,{0},0,0,{0},0,0,{0},0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,{0},0,0,{0},0,0,{0},0,0,{0},0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,{0},0,0,{0},0,0,{0},0,0,{0},0,0,{0}-0-0-0}-0-0}-0-0}-0-0}-0-0}}}}}

Copy:

**Spoiler:**

And so on.

3,{{0-0}-{0,0},0-{0,0}-0-0-0,0,0}

Follows the same story at the start.

3,{x-0,0,0} = 3,{{{{{{x-0,0}}}}} = 3,{{{{{{{{{x-0}}}}}{x-0,0}}}}

Reduce:

{x-0} = {x-0-0} = {x-0-0} = {x-{0,0}-0-0} = {{{{{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0}-0}-0}-0}-0}}}}}

Copy:

3,{{{{{{{{{x-0}}}}}{x-0,0}}}} = 3,{{{{{{{{{{{{{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0}-0}-0}-0}-0}}}}}}}}}{x-0,0},{{{{{{{{{{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0}-0}-0}-0}-0}}}}}}}}}{x-0,0},{{{{{{{{{{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0}-0}-0}-0}-0}}}}}}}}}{x-0,0},{{{{{{{{{{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0}-0}-0}-0}-0}}}}}}}}}{x-0,0},{{{{{{{{{{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0}-0}-0}-0}-0}}}}}}}}}{x-0,0}}}}

Reduce:

{x-0,0,0,0,{0}-0-0} = {{{{{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-0-0}-0}-0}-0}-0}}}}}

Copy:

3,{{{{{{{{{{{{{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0}-0}-0}-0}-0}}}}}}}}}{x-0,0},{{{{{{{{{{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0}-0}-0}-0}-0}}}}}}}}}{x-0,0},{{{{{{{{{{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0}-0}-0}-0}-0}}}}}}}}}{x-0,0},{{{{{{{{{{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0}-0}-0}-0}-0}}}}}}}}}{x-0,0},{{{{{{{{{{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0}-0}-0}-0}-0}}}}}}}}}{x-0,0}}}} = 3,{{{{{{{{{{{{{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0}-0}-0}-0}-0}}}}}}}}}{x-0,0},{{{{{{{{{{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0}-0}-0}-0}-0}}}}}}}}}{x-0,0},{{{{{{{{{{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0}-0}-0}-0}-0}}}}}}}}}{x-0,0},{{{{{{{{{{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-0-0}-0}-0}-0}-0}}}}}}}}}{x-0,0},{{{{{{{{{{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{{{{{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-0-0}-0}-0}-0}-0}}}}}-0}-0}-0}-0}}}}}}}}}{x-0,0},{{{{{{{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{{{{{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-0-0}-0}-0}-0}-0}}}}}-0}-0}-0}-0}}}}}}}}}{x-0,0},{{{{{{{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{{{{{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-0-0}-0}-0}-0}-0}}}}}-0}-0}-0}-0}}}}}}}}}{x-0,0},{{{{{{{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{{{{{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-0-0}-0}-0}-0}-0}}}}}-0}-0}-0}-0}}}}}}}}}{x-0,0},{{{{{{{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{x-0,0,0,0,{0}-{{{{{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-{x-0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0},0,0,0,{0}-0-0}-0}-0}-0}-0}}}}}-0}-0}-0}-0}}}}}}}}}{x-0,0}}}}

...

That is actually quite weak

And that's why I'd rather treat my notation as ordinals and enumerate their fixed points, if possible. I.e. I should be able to make my notation behave in the same way that φ(1@ω), φ(1@ω+1), φ(1@ω2), φ(1@ω^2)... behaves.

### Re: Your number is, in fact, not bigger!

Hmm, based on your examples I'm afraid I don't think the notation will reach the SVO.

More analysis:

{0-0} = z_0

{0-0,0} = z_1

{0-0,0,0} = z_2

{0-{0}} = z_ω

{0-{0-0}} = z_z_0

{0,0-0} = phi(3,0)

{0,0-0,0} = z_(phi(3,0)+1)

{0,0-0,0,0} = z_(phi(3,0)+2)

(0,0-{0}} = z_(phi(3,0)+ω}

{0,0-{0-0}} = z_(phi(3,0)+z_0)

{0,0-{0,0-0}} = z_(phi(3,0) + phi(3,0))

{0,0-{0,0-{0,0-0}}} = z_z_(phi(3,0) 2)

{0,0,0-0} = phi(3,1)

{0,0,0,0-0} = phi(3,2)

{{0}-0} = phi(3,ω)

{{0-0}-0} = phi(3,z_0)

{{{0-0}-0}-0} = phi(3, phi(3, z_0))

{0-0-0} = phi(4,0)

{0-0-0,0} = z_(phi(4,0)+1)

{0-0-{0-0-0}} = z_(phi(4,0) 2)

{0-0-{0-0-{0-0-0}}} = z_z_(phi(4,0) 2)

{0-0,0-0} = phi(3,phi(4,0)+1)

{0-0,0,0-0} = phi(3,phi(4,0)+2)

{0-{0}-0} = phi(3,phi(4,0)+ω)

{0-{0-0-0}-0} = phi(3, phi(4,0) 2)

{0-{0-{0-0-0}-0}-0} = phi(3, phi(3, phi(4,0) 2))

{0,0-0-0} = phi(4,1)

{0,0,0-0-0} = phi(4,2)

{{0}-0-0} = phi(4,ω)

{{0-0-0}-0-0} = phi(4, phi(4,0))

{{{0-0-0}-0-0}-0-0} = phi(4, phi(4, phi(4,0))

{0-0-0-0} = phi(5,0)

so this is clearly headed to phi(ω,0) not the SVO.

[/quote]

Unfortunately, it is not sufficient to simply define a notation as an ordinal - you must also say how the notation reduces. Don't give up, there's definitely ways to do this.

More analysis:

{0-0} = z_0

{0-0,0} = z_1

{0-0,0,0} = z_2

{0-{0}} = z_ω

{0-{0-0}} = z_z_0

{0,0-0} = phi(3,0)

{0,0-0,0} = z_(phi(3,0)+1)

{0,0-0,0,0} = z_(phi(3,0)+2)

(0,0-{0}} = z_(phi(3,0)+ω}

{0,0-{0-0}} = z_(phi(3,0)+z_0)

{0,0-{0,0-0}} = z_(phi(3,0) + phi(3,0))

{0,0-{0,0-{0,0-0}}} = z_z_(phi(3,0) 2)

{0,0,0-0} = phi(3,1)

{0,0,0,0-0} = phi(3,2)

{{0}-0} = phi(3,ω)

{{0-0}-0} = phi(3,z_0)

{{{0-0}-0}-0} = phi(3, phi(3, z_0))

{0-0-0} = phi(4,0)

{0-0-0,0} = z_(phi(4,0)+1)

{0-0-{0-0-0}} = z_(phi(4,0) 2)

{0-0-{0-0-{0-0-0}}} = z_z_(phi(4,0) 2)

{0-0,0-0} = phi(3,phi(4,0)+1)

{0-0,0,0-0} = phi(3,phi(4,0)+2)

{0-{0}-0} = phi(3,phi(4,0)+ω)

{0-{0-0-0}-0} = phi(3, phi(4,0) 2)

{0-{0-{0-0-0}-0}-0} = phi(3, phi(3, phi(4,0) 2))

{0,0-0-0} = phi(4,1)

{0,0,0-0-0} = phi(4,2)

{{0}-0-0} = phi(4,ω)

{{0-0-0}-0-0} = phi(4, phi(4,0))

{{{0-0-0}-0-0}-0-0} = phi(4, phi(4, phi(4,0))

{0-0-0-0} = phi(5,0)

so this is clearly headed to phi(ω,0) not the SVO.

And that's why I'd rather treat my notation as ordinals and enumerate their fixed points, if possible. I.e. I should be able to make my notation behave in the same way that φ(1@ω), φ(1@ω+1), φ(1@ω2), φ(1@ω^2)... behaves.

[/quote]

Unfortunately, it is not sufficient to simply define a notation as an ordinal - you must also say how the notation reduces. Don't give up, there's definitely ways to do this.

- Vytron
**Posts:**429**Joined:**Mon Oct 19, 2009 10:11 am UTC**Location:**The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P

### Re: Your number is, in fact, not bigger!

Deedlit wrote:{0,0-0,0} = z_(phi(3,0)+1)

So in your analysis, what value is this?:

{0-{0,0-0},0}

### Re: Your number is, in fact, not bigger!

with three arguments, each increment of the third argument increments to the next z-ordinal, each increment of the second argument increments to the next phi(3,x) ordinal, and each increment of the first argument increments to the next phi(4,x) ordinal.

Since {0,0-0} = phi(3,0), {0-{0,0-0},0} = phi(3, phi(4,0) + phi(3,0))

Since {0,0-0} = phi(3,0), {0-{0,0-0},0} = phi(3, phi(4,0) + phi(3,0))

- Vytron
**Posts:**429**Joined:**Mon Oct 19, 2009 10:11 am UTC**Location:**The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P

### Re: Your number is, in fact, not bigger!

But {0-{0,0-0},0} happens before {0,0-0,0}, i.e. {0,0-0} has to go over all these progressions:

{0,0-0},0

{0,0-0},{0,0-0}

0,{0,0-0} = {0,0-0}<,{0,0-0}>

<{0->0<}>{0,0-0} = {{0,0-0},0}

{<{0->0<}>}{{0,0-0}} = {{{0,0-0}},0}

{0-{0,0-0}} = <{>0,0-0<}>

{0-{0,0-0}},0

{0-{0,0-0}},{0-{0,0-0}}

0,{0-{0,0-0}} = {0-{0,0-0}}<,{0-{0,0-0}}>

<{>0,0-0<}>{0-{0,0-0}} = {{0-{0,0-0}},0}

{<{>0,0-0<}>}{{0-{0,0-0}}} = {{0-{0,0-0}},0}

{0-{0,0-0},0} = <{>0-{0,0-0}<}>

And:

{0,0-0,0} = <{0->{0,0-0}<}>

So {0,0-0,0} can't be z_(phi(3,0)+1) because {0,0-0,0} > {0-{0,0-0},0}, and {0-0-0} can't be phi(4,0) because {0-0-0} > {0,0-0,0} > {0-{0,0-0},0} (so if the last term is phi(3, phi(4,0) + phi(3,0)) the previous two terms should be higher.)

{0,0-0},0

{0,0-0},{0,0-0}

0,{0,0-0} = {0,0-0}<,{0,0-0}>

<{0->0<}>{0,0-0} = {{0,0-0},0}

{<{0->0<}>}{{0,0-0}} = {{{0,0-0}},0}

{0-{0,0-0}} = <{>0,0-0<}>

{0-{0,0-0}},0

{0-{0,0-0}},{0-{0,0-0}}

0,{0-{0,0-0}} = {0-{0,0-0}}<,{0-{0,0-0}}>

<{>0,0-0<}>{0-{0,0-0}} = {{0-{0,0-0}},0}

{<{>0,0-0<}>}{{0-{0,0-0}}} = {{0-{0,0-0}},0}

{0-{0,0-0},0} = <{>0-{0,0-0}<}>

And:

{0,0-0,0} = <{0->{0,0-0}<}>

So {0,0-0,0} can't be z_(phi(3,0)+1) because {0,0-0,0} > {0-{0,0-0},0}, and {0-0-0} can't be phi(4,0) because {0-0-0} > {0,0-0,0} > {0-{0,0-0},0} (so if the last term is phi(3, phi(4,0) + phi(3,0)) the previous two terms should be higher.)

### Re: Your number is, in fact, not bigger!

Sorry, I gave you the value of {0-{0,0-0}-0}.

{0-{0,0-0}} is a weird expression, it's bigger than {0,0-0} but not much bigger; {0,0-0} resolves to <{0->{0}<}>, and {0-{0,0-0}} just has an extra layer of {0-x} to it, so it has the same limit. So {0-{0,0-0}} is also at phi(3,0)

So {0-{0,0-0},0} is at z_(phi(3,0)+1)

{0-{0,0-0}} is a weird expression, it's bigger than {0,0-0} but not much bigger; {0,0-0} resolves to <{0->{0}<}>, and {0-{0,0-0}} just has an extra layer of {0-x} to it, so it has the same limit. So {0-{0,0-0}} is also at phi(3,0)

So {0-{0,0-0},0} is at z_(phi(3,0)+1)

- Vytron
**Posts:**429**Joined:**Mon Oct 19, 2009 10:11 am UTC**Location:**The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P

### Re: Your number is, in fact, not bigger!

I don't think that can be right, if:

{0,0-0} = phi(3,0)

Then:

{0,0-0},0 = phi(3,0)+1

{0,0-0},0,0 = phi(3,0)+2

{0,0-0},{0} = phi(3,0)+ω

{0,0-0},{0,0-0} = phi(3,0)*2

0,{0,0-0} = phi(3,0)*ω

{0-0},{0,0-0} = phi(3,0)*phi(2,0)

{{0,0-0},0} = phi(3,0)^2

{{{0,0-0}}} = ε_phi(3,0)

{{{{0,0-0}}}} = ε_ε_phi(3,0)

{0-{0,0-0}} = ζ_phi(3,0)

{{0-{0,0-0}}} = ε_ζ_phi(3,0)

{{{0-{0,0-0}}}} = ε_ε_ζ_phi(3,0)

{0-{0,0-0},0} = ζ_{phi(3,0)+1}

{0-{0-{0,0-0},0},0} = ζ_ζ_{phi(3,0)+1}

{0,0-0,0} = phi(3,1)

{0,0-{0}} = phi(3,ω)

{0,0-{0,0-0}} = phi(3,phi(3,0))

{0,0-{0,0-{0,0-0}}} = phi(3,phi(3,phi(3,0)))

{0,0,0-0} = phi(4,0)

I.e. You say {0-{0,0-0}} "adds an extra layer", however, I have defined "adding an extra layer" as n+1,{0,0-0}, and n,{0,0-0},0 is <(>n<),{0,0-0}> so it already is over phi(3,0), and it's a long way from {0-{0,0-0},0}.

Of note:

I already had this discussion with WarDaft it begins here, but here's the replay:

I think the mistake is having:

n,{0,0-0} = n,<{0->0<}>

And then thinking:

n,{0-{0,0-0}} = n,{0-<{0->0<}>}

Which is not the case, we don't reduce n,{0-X}, We go to <{>X<}>.

{0,0-0} = phi(3,0)

Then:

{0,0-0},0 = phi(3,0)+1

{0,0-0},0,0 = phi(3,0)+2

{0,0-0},{0} = phi(3,0)+ω

{0,0-0},{0,0-0} = phi(3,0)*2

0,{0,0-0} = phi(3,0)*ω

{0-0},{0,0-0} = phi(3,0)*phi(2,0)

{{0,0-0},0} = phi(3,0)^2

{{{0,0-0}}} = ε_phi(3,0)

{{{{0,0-0}}}} = ε_ε_phi(3,0)

{0-{0,0-0}} = ζ_phi(3,0)

{{0-{0,0-0}}} = ε_ζ_phi(3,0)

{{{0-{0,0-0}}}} = ε_ε_ζ_phi(3,0)

{0-{0,0-0},0} = ζ_{phi(3,0)+1}

{0-{0-{0,0-0},0},0} = ζ_ζ_{phi(3,0)+1}

{0,0-0,0} = phi(3,1)

{0,0-{0}} = phi(3,ω)

{0,0-{0,0-0}} = phi(3,phi(3,0))

{0,0-{0,0-{0,0-0}}} = phi(3,phi(3,phi(3,0)))

{0,0,0-0} = phi(4,0)

I.e. You say {0-{0,0-0}} "adds an extra layer", however, I have defined "adding an extra layer" as n+1,{0,0-0}, and n,{0,0-0},0 is <(>n<),{0,0-0}> so it already is over phi(3,0), and it's a long way from {0-{0,0-0},0}.

Of note:

I already had this discussion with WarDaft it begins here, but here's the replay:

Vytron wrote:Okay, so I have here:

n[0-0-[0-0-0]] = n[0-0-[~[1-0]-0~]] = Γ_Γ_0

n[0-0-[0-0-[0-0-0]]] = n[0-0-[0-0-[~[1-0]-0~]]] = Γ_Γ_Γ_0

n[0-0-[0-0-[0-0-[0-0-0]]]] = n[0-0-[0-0-[0-0-[~[1-0]-0~]]]] = Γ_Γ_Γ_Γ_0

n[0-0,0-0] = n[0-0-[~0-0-[0-0-0]~]]

(2[0-0,0-0] should match WarDaft)

Send

4[0-0-[0-0,0-0]]

For f_{Γ_(first fixed point of a -> Γa)}

WarDaft wrote:That's not how ordinals work. It's a fixed point, the point where δ = Γδ. So Γδ is still just δ. If your notation works so that you can truly substitute ordinals like that, then that should be equivalent to 5[0-0,0-0].

Vytron wrote:Well, I think I can do away with FGH comparisons and beat you old style:

n[0-0,0-0] = f_{Γ_Γ_...Γ_Γ_0}(n)

n[0-0,0-0],0 = n[0-0,0-0] n[0-0,0-0] ... n[0-0,0-0]

n[0-0,0-0],0,0 = n[0-0,0-0],0 n[0-0,0-0],0 ... n[0-0,0-0],0

n[0-0,0-0],0,0,0 = n[0-0,0-0],0,0 n[0-0,0-0],0,0 ... n[0-0,0-0],0,0

n[0-0,0-0],1 = n[0-0,0-0],0,~

n[0-0,0-0],1,1 = n[0-0,0-0],1,0,~

n[0-0,0-0],0,1 = n[0-0,0-0],1,~

n[0-0,0-0],0,0,1 = n[0-0,0-0],0,1,~~

n[0-0,0-0],0,0,0,1 = n[0-0,0-0],0,0,1,~~~

n[0-0,0-0],2 = n[0-0,0-0],0,~1

n[0-0,0-0],3 = n[0-0,0-0],0,~1,2

n[0-0,0-0],[0] = n[0-0,0-0],n

n[0-0,0-0],[0],2 = n[0-0,0-0],[0],0,~1

n[0-0,0-0],[0],3 = n[0-0,0-0],[0],0,~1,2

n[0-0,0-0],[0],[0] = n[0-0,0-0],[0],n

n[0-0,0-0],0,[0] = n[0-0,0-0],[0],~

n[0-0,0-0],0,0,[0] = n[0-0,0-0],0,[0],~~

n[0-0,0-0],1,[0] = n[0-0,0-0],0,~[0]

n[0-0,0-0],2,[0] = n[0-0,0-0],0,~1,[0]

n[0-0,0-0],[0]~[0] = n[0-0,0-0],n,[0]

n[0-0,0-0],[0,0] = n[0-0,0-0],[0]~[0]~

n[0-0,0-0],[0,0,0] = n[0-0,0-0],[0,0]~[0,0]~

n[0-0,0-0],[1] = n[0-0,0-0],[0,~]

n[0-0,0-0],[2] = n[0-0,0-0],[0,~1]

n[0-0,0-0],[[0]] = n[0-0,0-0],[n]

n[0-0,0-0],[[[0]]] = n[0-0,0-0],[[n]]

n[0-0,0-0],[0-0] = n[0-0,0-0],[~[0]~]

n[0-0,0-0],[0-0,0] = n[0-0,0-0],[~[0-0],0~]

n[0-0,0-0],[0,0-0] = n[0-0,0-0],[0-[~0-0]~]

n[0-0,0-0],[0-0-0] = n[0-0,0-0],[~[1-0]-0~]

n[0-0,0-0],[0-0-0,0] = n[0-0,0-0],[[~[1-0]-0~]-[0-0-0],0]

n[0-0,0-0],[0-0-[0-0-0]] = n[0-0,0-0],[0-0-[~[1-0]-0~]]

n[0-0,0-0],[0-0,0-0] = n[0-0,0-0],[0-0-[~0-0-[0-0-0]~]]

n[0-0,0-0],[0-0,0-0],[0-0,0-0] = n[0-0,0-0],[0-0,0-0],[0-0-[~0-0-[0-0-0]~]]

n,0,[0-0,0-0] = n[0-0,0-0],~

n,0,0,[0-0,0-0] = n,0,[0-0,0-0],~~

n,1,[0-0,0-0] = n,0,~[0-0,0-0]

n,2,[0-0,0-0] = n,0,~1,[0-0,0-0]

n[0],[0-0,0-0] = n,n,[0-0,0-0]

n,0,[0],[0-0,0-0] = n[0],~[0-0,0-0]

n[0]~[0],[0-0,0-0] = n,n,[0],~[0-0,0-0]

n[0,0],[0-0,0-0] = n[0]~[0]~,[0-0,0-0]

n[0,0,0],[0-0,0-0] = n[0,0]~[0,0]~,[0-0,0-0]

n[1],[0-0,0-0] = n[0,~],[0-0,0-0]

n[2],[0-0,0-0] = n[0,~1],[0-0,0-0]

n[[0]],[0-0,0-0] = n[n],[0-0,0-0]

n[[[0]]],[0-0,0-0] = n[[n]],[0-0,0-0]

n[0-0],[0-0,0-0] = n[~[0]~],[0-0,0-0]

n[0-0,0],[0-0,0-0] = n[~[0-0],0~],[0-0,0-0]

n[0,0-0],[0-0,0-0] = n[0-[~0-0]~],[0-0,0-0]

n[0-0-0],[0-0,0-0] = n[~[1-0]-0~],[0-0,0-0]

n[0-0-0,0],[0-0,0-0] = n[[~[1-0]-0~]-[0-0-0],0],[0-0,0-0]

n[0-0-[0-0-0]],[0-0,0-0] = n[0-0-[~[1-0]-0~]],[0-0,0-0]

n[0-0,0-0]~[0-0,0-0] = n[0-0-[~0-0-[0-0-0]~],[0-0,0-0]

n[0-0,0-0]~[0-0,0-0]~[0-0,0-0] = n[0-0-[~0-0-[0-0-0]~]~[0-0,0-0]

n[[0-0,0-0]] = n[0-0,0-0]~[0-0,0-0]~

(i.e. n[0-0,0-0]~[0-0,0-0]~ is n[0-0,0-0]~[0-0,0-0]...~[0-0,0-0] with n ~s)

I bolded the important parts.

Now, to get to n[[0-0,0-0]] (which is still far away from n[0-0-[0-0,0-0]]) one had to undergo limits of the [0-0-[~0-0-[0-0-0]~] kind several times, so how could 4[0-0-[0-0,0-0]] equal 5[0-0,0-0]?

WarDaft wrote:Alright, I'll accept that.

Vytron wrote:(Also, I thought about a second argument to show how 4[0-0-[0-0,0-0]] can't be 5[0-0,0-0] - At the very base of the continuation of my progression, I have 4[0-0-[0-0,0-0]],0 = n[0-0-[0-0,0-0]] n[0-0-[0-0,0-0]] n[0-0-[0-0,0-0]] n[0-0-[0-0,0-0]] 4[0-0-[0-0,0-0]] i.e. there's nesting applied to the previous function. It means 2[0-0-[0-0,0-0]],0 is (((2[0-0-[0-0,0-0]])[0-0-[0-0,0-0]])[0-0-[0-0,0-0]]) which is already much larger than 5[0-0-[0-0,0-0]])

I think the mistake is having:

n,{0,0-0} = n,<{0->0<}>

And then thinking:

n,{0-{0,0-0}} = n,{0-<{0->0<}>}

Which is not the case, we don't reduce n,{0-X}, We go to <{>X<}>.

### Re: Your number is, in fact, not bigger!

Don't take my accepting of that as meaning it was true, merely that I accepted it without further fuss.

In my own notation, (((0,1))) is larger than ((0,1),0), even though the latter is applying a more powerful function [ ( ,0) instead of (( ))] to the same base value. The difference is in what is reduced. In the first case, only the outer structure is reduced, to ((0,1)), in the second case, the (0,1) is also reduced at the same time, this means it is behaving differently and has a lesser value than you might expect.

If to reduce {0-{0,0-0}} to the next step, you reduce the {0,0-0} component, then it may not behave as you expect.

In my own notation, (((0,1))) is larger than ((0,1),0), even though the latter is applying a more powerful function [ ( ,0) instead of (( ))] to the same base value. The difference is in what is reduced. In the first case, only the outer structure is reduced, to ((0,1)), in the second case, the (0,1) is also reduced at the same time, this means it is behaving differently and has a lesser value than you might expect.

If to reduce {0-{0,0-0}} to the next step, you reduce the {0,0-0} component, then it may not behave as you expect.

All Shadow priest spells that deal Fire damage now appear green.

Big freaky cereal boxes of death.

### Re: Your number is, in fact, not bigger!

Vytron wrote:I don't think that can be right, if:

{0,0-0} = phi(3,0)

Then:

{0,0-0},0 = phi(3,0)+1

{0,0-0},0,0 = phi(3,0)+2

{0,0-0},{0} = phi(3,0)+ω

{0,0-0},{0,0-0} = phi(3,0)*2

0,{0,0-0} = phi(3,0)*ω

{0-0},{0,0-0} = phi(3,0)*phi(2,0)

{{0,0-0},0} = phi(3,0)^2

{{{0,0-0}}} = ε_phi(3,0)

{{{{0,0-0}}}} = ε_ε_phi(3,0)

{0-{0,0-0}} = ζ_phi(3,0)

{{0-{0,0-0}}} = ε_ζ_phi(3,0)

{{{0-{0,0-0}}}} = ε_ε_ζ_phi(3,0)

{0-{0,0-0},0} = ζ_{phi(3,0)+1}

{0-{0-{0,0-0},0},0} = ζ_ζ_{phi(3,0)+1}

{0,0-0,0} = phi(3,1)

{0,0-{0}} = phi(3,ω)

{0,0-{0,0-0}} = phi(3,phi(3,0))

{0,0-{0,0-{0,0-0}}} = phi(3,phi(3,phi(3,0)))

{0,0,0-0} = phi(4,0)

You could say that {0-{0,0-0}} = ζ_phi(3,0), as ζ_phi(3,0) = phi(3,0). However, it looks like you are changing the definition of {0,0-0,0}. From your previous post, you stated that if the rightmost element is larger than 0, we ignore everything previous and just apply the previous rule. So {0,0-0,0} = <{>{0,0-0}<}>, which is ζ_{phi(3,0)+1}, and it does not dominate notations like {0-{0-{0,0-0},0},0}, as it cannot reduce to them.

I.e. You say {0-{0,0-0}} "adds an extra layer", however, I have defined "adding an extra layer" as n+1,{0,0-0}, and n,{0,0-0},0 is <(>n<),{0,0-0}> so it already is over phi(3,0), and it's a long way from {0-{0,0-0},0}.

You are taking the phrase "adds an extra layer" out of context. What I meant was, while {0,0-0} reduced to {0-{0}}, {0-{0-{0]}}, {0-{0-{0-{0}}}, etc., {0-{0,0-0}} reduces to {{0-{0-{0}}}}, {{{0-{0-{0-{0}}}}}}, {{{{0-{0-{0-{0-{0}}}}}}}}, etc. This is what we discussed before, although for the same n {0-{0,0-0}} reduces to a notation with a larger ordinal than what {0,0-0} reduces to for the same n, the limit of those ordinals is the same. So {0-{0,0-0}} cannot correspond to a higher ordinal than phi(3,0) as all the notations it reduces to have ordinals less than phi(3.0).

I already had this discussion with WarDaft it begins here, but here's the replay:

You may have beat WarDaft into submission here.

I think the mistake is having:

n,{0,0-0} = n,<{0->0<}>

And then thinking:

n,{0-{0,0-0}} = n,{0-<{0->0<}>}

Which is not the case, we don't reduce n,{0-X}, We go to <{>X<}>.

It doesn't matter - any combination of X -> <{>X<}> and/or X -> <{0-><X}> will not take a notation below {0,0-0} to a notation above {0,0-0}, which is how {0,0-0} got to be phi(3,0) in the first place. Since {0-{0,0-0}} always resolves to a notation below {0,0-0}, it cannot have a larger ordinal.

### Re: Your number is, in fact, not bigger!

Vytron: Why would i plant forests on clouds for them to come crashing down? i'd rather perfect the root of my notation endlessly no matter how much power i lose or gain so long as i end up getting absolute confidence in my FGH comparisons, only when i have a solid grasp on my own claims i should shoot for further ones. Like you i have many ideas for way higher complicated extensions but if i can't work out the kinks in the base i have no business flexing those.

On a positive note. i found it easy to not mess up with ordinal exponent laws if i just replace ω with 3 in hypercalc and compare my intended results. For low towers like ω^ω^ω^ω this is feasible.

General sizing GFDQM

Nesting level: 1

¿c¡e¡0!d!b?a

How +1 on a variable affects the fgh comparison

# stands for "ordinal"

a: f#(n+1)

b: f#+1(n)

c: f#+#(n)

d: f#*# (not #+ω or #*ω because +1 on c and d respectively is the limit of chaining the separator as a whole, which is #+# at each link)

e: f#^ω

The limit of this is

¿c¡0¡0!0¡0!0!b?a

repetitively doing (...(#)^ω)^ω...) leads to #^ω^ω

Since we were at ω^ω^ω, we arrive at (ω^ω^ω)^(ω^ω) = ω^(ω^ω)2 (i mistakenly went from ω^ω^ω to ω^ω^ω^2 at this step in my previous comparisons, in the heat of the moment i gave d the power of e rendering all old comparisons *wrong*)

¿c¡f¡0!e¡0!d!b?a

How +1 on a variable affects the fgh comparison

# stands for "ordinal"

a: f#(n+1)

b: f#+1(n)

c: f#+#(n)

d: f#*#(n)

e: f#^ω

f: f#^ω^ω

repetitively doing (...(#)^ω^ω)...^ω^ω) will lead to ω^ω^ω in the limit

¿c¡g¡0!f¡0!e¡0!d!b?a

Since we began at ω^(ω^ω)2, we arrive at (ω^(ω^ω)2)^(ω^ω^ω) > ω^ω^(ω^ω+ω) > ω^ω^ω^ω as the starting point

g: #^ω^ω^ω , limiting at ^ω^ω^ω^ω gives a low bound at ω^ω^(ω^ω^ω+ω^ω) > ω^ω^ω^ω^ω the limit a.k.a another ¡0! in the chain

I'm confident ¿0¡0¡1!0!0?n > fε

Continuing. effect of +1 on variables

¿c¡e¡1!d!b?a

a: f#(n+1)

b: f#+1(n)

c: f#+#(n)

d: f#*#(n)

The limit of d is not e+1 but actually to become of 0¡0!0 form

¿0¡0¡1!0¡0!0!0? > fε0^ω

Continuing

¿c¡f¡1!e¡0!d!b?a

a: f#(n+1)

b: f#+1(n)

c: f#+#(n)

d: f#*#(n)

e: f#^ω

Limit of e being

¿b¡0¡1!0¡0!0¡0!0!c?a > fε0^ω^ω

Continues the 0<¡0!0> progression as above

So back here

¿c¡e¡1!d!b?a

we get

a: f#(n+1)

b: f#+1(n)

c: f#+#(n)

d: f#*#(n)

e: f#^ε0(n)

The limit of that is for e to become of 0¡0!0 form itself. and (...ε0)^ε0)...^ε0 should cap at ε0^ω

¿0¡0¡0!0¡1!0!0?10 > fε0^ε0^ω(10)

Continuing

¿c¡f¡0!e¡1!d!b?a >

f: f#^ε0^ω(n)

¿0¡0¡0!0¡0!0¡1!0!0?10 > fε0^ε0^ω^ω(10)

¿0¡0¡1!0¡1!0!0?10 > fε0^ε0^ε0(10)

Implying ¿0¡0¡2!0!0? > fε1

...to be continued

On a positive note. i found it easy to not mess up with ordinal exponent laws if i just replace ω with 3 in hypercalc and compare my intended results. For low towers like ω^ω^ω^ω this is feasible.

General sizing GFDQM

Nesting level: 1

¿c¡e¡0!d!b?a

How +1 on a variable affects the fgh comparison

# stands for "ordinal"

a: f#(n+1)

b: f#+1(n)

c: f#+#(n)

d: f#*# (not #+ω or #*ω because +1 on c and d respectively is the limit of chaining the separator as a whole, which is #+# at each link)

e: f#^ω

The limit of this is

¿c¡0¡0!0¡0!0!b?a

repetitively doing (...(#)^ω)^ω...) leads to #^ω^ω

Since we were at ω^ω^ω, we arrive at (ω^ω^ω)^(ω^ω) = ω^(ω^ω)2 (i mistakenly went from ω^ω^ω to ω^ω^ω^2 at this step in my previous comparisons, in the heat of the moment i gave d the power of e rendering all old comparisons *wrong*)

¿c¡f¡0!e¡0!d!b?a

How +1 on a variable affects the fgh comparison

# stands for "ordinal"

a: f#(n+1)

b: f#+1(n)

c: f#+#(n)

d: f#*#(n)

e: f#^ω

f: f#^ω^ω

repetitively doing (...(#)^ω^ω)...^ω^ω) will lead to ω^ω^ω in the limit

¿c¡g¡0!f¡0!e¡0!d!b?a

Since we began at ω^(ω^ω)2, we arrive at (ω^(ω^ω)2)^(ω^ω^ω) > ω^ω^(ω^ω+ω) > ω^ω^ω^ω as the starting point

g: #^ω^ω^ω , limiting at ^ω^ω^ω^ω gives a low bound at ω^ω^(ω^ω^ω+ω^ω) > ω^ω^ω^ω^ω the limit a.k.a another ¡0! in the chain

I'm confident ¿0¡0¡1!0!0?n > fε

_{0}(n+1) holdsContinuing. effect of +1 on variables

¿c¡e¡1!d!b?a

a: f#(n+1)

b: f#+1(n)

c: f#+#(n)

d: f#*#(n)

The limit of d is not e+1 but actually to become of 0¡0!0 form

¿0¡0¡1!0¡0!0!0? > fε0^ω

Continuing

¿c¡f¡1!e¡0!d!b?a

a: f#(n+1)

b: f#+1(n)

c: f#+#(n)

d: f#*#(n)

e: f#^ω

Limit of e being

¿b¡0¡1!0¡0!0¡0!0!c?a > fε0^ω^ω

Continues the 0<¡0!0> progression as above

So back here

¿c¡e¡1!d!b?a

we get

a: f#(n+1)

b: f#+1(n)

c: f#+#(n)

d: f#*#(n)

e: f#^ε0(n)

The limit of that is for e to become of 0¡0!0 form itself. and (...ε0)^ε0)...^ε0 should cap at ε0^ω

¿0¡0¡0!0¡1!0!0?10 > fε0^ε0^ω(10)

Continuing

¿c¡f¡0!e¡1!d!b?a >

f: f#^ε0^ω(n)

¿0¡0¡0!0¡0!0¡1!0!0?10 > fε0^ε0^ω^ω(10)

¿0¡0¡1!0¡1!0!0?10 > fε0^ε0^ε0(10)

Implying ¿0¡0¡2!0!0? > fε1

...to be continued

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