Your number is, in fact, not bigger!

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

Postby Daggoth » Fri Mar 06, 2015 5:16 pm UTC

¿c¡e¡2¡d!b?a
How +1 on a variable affects the FGH comparison
a: f#(n+1)
b: f#+1(n)
c: f#+#(n)
d: f#*#(n)
¿c¡e¡2¡0¡0!0!b?a > ε1^ω
as shown the 0<¡0!0> progression limits at
¿c¡e¡2¡0¡1!0!b?a > ε1^ε0
¿c¡e¡2¡g¡1!f!b?a > ε1^ε0
here f: f#^ω can itself go into 0<¡0!0> for a limit of
¿c¡e¡2¡1¡1!0!b?a > ε1^ε0^ε0
so the above g: f#^ε0
g becoming 0¡0!0 hits the limit (...(ε0^ε0)^ε0...)^ε0, or #^ε0^ω
¿c¡e¡2¡0¡1!0¡1!0!b?a > ε1^ε0^ε0
back here ¿c¡e¡2¡d!b?a
e: f#^ε1
so ¿c¡0¡0!0¡2!d!b?a > ε1^ε1^ω
here
¿c¡f¡0!e¡2!d!b?a
e: #^ε1
f: #^ε1^ω
so ¿c¡0¡0!0¡0!0¡2!d!b?a > ε1^ε1^ω^ω
and
¿c¡0¡1!0¡2!d!b?a > ε1^ε1^ε0
¿c¡0¡2!0¡2!d!b?a > ε1^ε1^ε1
Thus ¿0¡0¡3!0!0?n > fε2(n+1)
This progression points to ¿0¡0¡0¡0!0!0!0?n > fεω(n-1)

here ¿c¡e¡0¡0!0!d!b?a
a-d behave exactly| as before with d: f#*# except that since we are at 2 level nesting, d is allowed to be 1 level or 0¡0¡0!0!0 before the e+1 hits.
more on that later.

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

Postby Vytron » Sat Mar 07, 2015 2:21 am UTC

Deedlit wrote: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.


No, it doesn't. I'll use the same argument against you that I used with WarDaft:

We have:

3{0,0-0} = 3{0-{0-{0-{0}}}

And:

4{0,0-0} = 4{0-{0-{0-{0-{0}}}}

Now, if what you said was true, then:

3{0-{0,0-0}} = 3{0-{0-{0-{0-{0}}}}

Making them both reach the same level.

However, to reach {0,0-0},0 one had to undergo:

<(>n<){0,0-0}>

That is, one had to reach:

4{0,0-0}
5{0,0-0}
6{0,0-0}
10{0,0-0}
10^10{0,0-0}
10^100{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}
(4,{0,0-0}){0,0-0}
(6,{0,0-0}){0,0-0}
(10,{0,0-0}){0,0-0}
(10^10,{0,0-0}){0,0-0}
(10^100,{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}){0,0-0}
((n,{0,0-0}),{0,0-0}){0,0-0}
<(>n<){0,0-0}> = n,{0,0-0},0

{0,0-0},0 dominates {0,0-0}, I have {0-{0,0-0}} eventually reducing to {0,0-0},0, so they both can't be at the same level, and 3{0-{0,0-0}} is not 3{0-{0-{0-{0-{0}}}} (the 0,0 doesn't reduce.)

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

Postby WarDaft » Sat Mar 07, 2015 2:45 am UTC

Vytron, what, exactly is the reduction for 3{0-{0,0-0}}, step by step? Just because {0-X} is a more powerful function in general does not mean it is always larger than X,0, for the same X in both cases.

If 3{0-{0,0-0}} does not become 3{0-{0-{0-{0-{0}}}}, what does it become?

And I've already said that my acceptance of your argument wasn't because you were right, but because I didn't feel like arguing the point.
Last edited by WarDaft on Sat Mar 07, 2015 3:01 am UTC, edited 1 time in total.
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Deedlit
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Re: Your number is, in fact, not bigger!

Postby Deedlit » Sat Mar 07, 2015 2:53 am UTC

All right, you're going to have to explain your rules again, because before the entire definition for the - notation was
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]>


and I supposed that by "Follows the rules of above", you meant that n,{0-x} would resolve the same way as n,x would, which would mean the x would indeed reduce. When I asked for an explanation of what "Follows the rules of above" meant, you answer seemed to agree with that. Certainly, if it resolves some other way, you haven't explained how it resolves.

Further, when I asked how certain examples resolved, you said in your answer:

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}}}}


so here, when the rightmost element that isn't 0 doesn't end in ,0, it _does_ resolve, according to your explanation.

Also, are you saying that {0-0,0-0,0} will eventually resolve to {0-{0,0-0}}?

To sum up, can you give the complete rules for your - notation?

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

Postby Vytron » Sat Mar 07, 2015 4:27 am UTC

I think it helps to see it upside down, i.e. lower lines reduce to previous lines, so bottom lines should dominate all previous ones. I.e.

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>

Should be pretty clear. Every lower line dominates a previous line.

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

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

And:

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

So:

3,{0-{0,0-0}} = 3,{{{{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}}}

Which should dominate n{0-{0-{0-{0-{0}}}}} or n{0-{0-{0-{0-{0-{0}}}}}}

Now, I'll claim this should go past f_φ(2,φ(3,0)+1)(3) in the FGH.
Last edited by Vytron on Sat Mar 07, 2015 4:36 am UTC, edited 1 time in total.

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

Postby Vytron » Sat Mar 07, 2015 4:33 am UTC

Also, are you saying that {0-0,0-0,0} will eventually resolve to {0-{0,0-0}}?


Yes.

{0-0,0-0,0,0} will also eventually resolve to all these values:

{X-0-{0-0,0-0,0}} (Where X is any value below {0-0,0-0,0,0})
{0-0-{0-0,0-0,0}}
{0-{0-0,0-0,0}}

And any other value {0-0,0-0,0} can sustain that has been defined before reaching {0-0,0-0,0,0}.

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

Postby Deedlit » Sat Mar 07, 2015 5:31 am UTC

I'm afraid I'm not seeing the general pattern here. For example,

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


doesn't seem to follow the pattern of any previous example.

I don't know what "Every lower line dominates a previous line" means.

It may well be that what you have in your mind can go up to the SVO, but you need to give a precise definition of it, with general rules rather than only examples.

{0-0,0-0,0,0} will also eventually resolve to all these values:

{X-0-{0-0,0-0,0}} (Where X is any value below {0-0,0-0,0,0})
{0-0-{0-0,0-0,0}}
{0-{0-0,0-0,0}}


wait, {0-0,0-0,0,0} can resolve to {X-0-{0-0,0-0,0}}? It definitely looks like the latter should be larger.


And any other value {0-0,0-0,0} can sustain that has been defined before reaching {0-0,0-0,0,0}.


You need to define an ordering on you notations for that to help.

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

Postby Vytron » Sat Mar 07, 2015 8:27 am UTC

Okay, here's some ordering.

There's these series (separated by spaces):

0 0,0 0,0,0 0,0,0,0 0,0,0,0,0...

When these series are at the rightmost point in:

n,X

You drop a 0 and return: <(>n<),X>

And then there's these series:

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

When these series are at the rightmost point in:

n,X

You transform the rightmost {0} into <,0>

There's:

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

When these series are at the rightmost point in:

n,X

You drop a 0, and return <,Y> where Y is an early member of the series (0,0,0,0,{0} becomes <0,0,0,{0}>)

When there's ambiguity in the series:

You apply these rules to string a, in n,x,a of the rules of the other post (i.e. x ends in the penultimate biggest element of X).

{0,0} Becomes 0<,0>{0} if it's the rightmost element, otherwise, all previous elements follow the same rules at the right of {0,0}

There exist the series

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

And the series:

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

And the series:

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

Which have known reduction rules for {0,0}X

There's the series:

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

When these series are at the rightmost point in:

n,X

You transform the rightmost {0,0} into <<,0>,{0}>

There's the series:

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

When these series are at the rightmost point in:

n,X

You drop a 0, and return <,Y> where Y is an early member of the series (0,0,0,0,{0,0} becomes <0,0,0,{0,0}>)

When there's ambiguity in the series:

You apply these rules to string a, in n,x,a of the rules of the other post (i.e. x ends in the penultimate biggest element of X).

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

There's the series:

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

On here you reduce X,{0,0}, create Y=X<,X>{0,0} And return Y<,Y>

I.e.:

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

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

{{0}} returns: {0<,0>}

{{0}} is an element bigger than any other {X} with no {}s on X, so all previous series follow the same rules on {{0}}X.

All other series also follow the same rules on X{{0}} (i.e. to know how {{0}} behaves just pretend {{0}} is a {0,0,0} with one 0 more than any terms in X).

The series:

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

If at the rightmost change the rightmost {{0}} to {0<,0>}

The series:

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

Drop a 0, and return <,Y> where Y is an early member of the series (0,0,0,0,{{0}} becomes <0,0,0,{{0}}>)

Series:

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

On here you reduce X,{{0}}, create Y=X<,X>{{0}} And return Y<,Y>

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

On here you reduce X,{{0}}, create Y=X<,X>{{0}} And return Y<,Y>

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

That is, to add a 0 to {X}, you have to have the element that created {X} in Y{X}

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

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

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

And so on.

{0,{0}} returns: {{0}<,{0}>}
{{0,0}} returns: {0<,0>{0}}
{{{0}}} returns: {{0<,0>}}

{0-0} returns: <{>0<}>

{0-0} is an element bigger than any other {X} with no -s on X, so all previous series follow the same rules on {0-0}X.

All other series also follow the same rules on X{0-0} (i.e. to know how {0-0} behaves just pretend {0-0} is a {{{0}}} with one {} nest more than any terms in X).

The series:

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

If at the rightmost change the rightmost {0-0} to <{>0<}>

The series:

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

Drop a 0, and return <,Y> where Y is an early member of the series (0,0,0,0,{0-0} becomes <0,0,0,{0-0}>)

Series:

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

On here you reduce X,{0-0}, create Y=X<,X>{0-0} And return Y<,Y>

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

On here you reduce X,{0-0}, create Y=X<,X>{0-0} And return Y<,Y>

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

On here you reduce X,{0-0}, create Y=X<,X>{0-0} And return Y<,Y>

{{0-0}} returns <{>0<}>,{0-0}
{{0-0},0} returns <{>0<}>,{0-0},{{0-0}}

That is, to add a 0 to {X}, you have to have the element that created {X} in Y{X}

{{0-0},0,0} returns <{>0<}>,{0-0},{{0-0},0}
{{0-0},0,0,0} returns <{>0<}>,{0-0},{{0-0},0},{{0-0},0,0}
{{0-0},0,0,0,0} returns <{>0<}>,{0-0},{{0-0},0},{{0-0},0,0},{{0-0},0,0,0}

{{0-0},{0}} returns {{0-0}<,0>}
{{0-0},{0},0} returns {{0-0}<,0>},{{0-0},{0}}
{{0-0},{0},0,0} returns {{0-0}<,0>},{{0-0},{0}},{{0-0},{0},0}

{{0-0},{0},{0}} returns {{0-0},{0}<,0>}

And so on.

{0,{0-0}} returns: {{0-0}<,{0-0}>}
{{{0-0}}} returns: {{<{>0<}>,{0-0}}}
{{{{0-0}}}} returns: {{{<{>0<}>,{0-0}}}}

{0-0,0} returns: <{>0-0<}>
{{0-0,0}} returns: <{>0-0<}>{0-0,0}
{{{0-0,0}}} returns: {<{>0-0<}>{0-0,0}}

{0-0,0,0} returns: <{>0-0,0<}>
{0-{0}} returns: <{>0-0<,0><}>
{0-{0},{0}} returns: <{>0-{0},0<,0><}>
{0-{0-0}} returns: <{>0-<{>0<}><}>
{0-{0-0,0}} returns: <{>0-<{>0-0<}><}>
{0-{0-0,0,0}} returns: <{>0-<{>0-0,0<}><}>
{0-{0-{0}}} returns: <{>0-<{>0-0<,0><}><}>
{0-{0-{0-0}}} returns: <{>0-<{>0-<{>0<}><}><}>
{0-{0-{0-0,0}}} returns: <{>0-<{>0-<{>0-0<}><}><}>
{0-{0-{0-0,0,0}}} returns: <{>0-<{>0-<{>0-0,0<}><}><}>
{0-{0-{0-{0}}}} returns: <{>0-<{>0-<{>0-0<,0><}><}><}>

And:

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

Now, please note that {0,0-0} never appeared in 0-X, so we use {0-{0,0-0}} to represent higher values.

{0,0-0} is an element bigger than any other {0-X} with no 0,0- on X, so all previous series follow the same rules on {0,0-0}X.

All other series also follow the same rules on X{0,0-0} (i.e. to know how {0,0-0} behaves just pretend {0,0-0} is a {0-{0-{0-{0}}}} with one - more than any terms in X).

The series:

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

If at the rightmost change the rightmost {0,0-0} to <{0->{0}<}>

The series:

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

Drop a 0, and return <,Y> where Y is an early member of the series (0,0,0,0,{0,0-0} becomes <0,0,0,{0,0-0}>)

Series:

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

On here you reduce X,{0,0-0}, create Y=X<,X>{0,0-0} And return Y<,Y>

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

On here you reduce X,{0,0-0}, create Y=X<,X>{0,0-0} And return Y<,Y>

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

On here you reduce X,{0,0-0}, create Y=X<,X>{0,0-0} And return Y<,Y>

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

On here you reduce X,{0,0-0}, create Y=X<,X>{0,0-0} And return Y<,Y>

{{0,0-0}} returns <{0->{0}<}>,{0,0-0}
{{0,0-0},0} returns <{0->{0}<}>,{0,0-0},{{0,0-0}}

That is, to add a 0 to {X}, you have to have the element that created {X} in Y{X}

{{0,0-0},0,0} returns <{0->{0}<}>,{0,0-0},{{0,0-0}},{{0,0-0},0}
{{0,0-0},0,0,0} returns <{0->{0}<}>,{0,0-0},{{0,0-0}},{{0,0-0},0},{{0-0},0,0}
{{0,0-0},0,0,0,0} returns <{0->{0}<}>,{0,0-0},{{0,0-0}},{{0,0-0},0},{{0-0},0,0},{{0-0},0,0,0}

{{0,0-0},{0}} returns {{0,0-0}<,0>}
{{0,0-0},{0},0} returns {{0,0-0}<,0>},{{0,0-0},{0}}
{{0,0-0},{0},0,0} returns {{0,0-0}<,0>},{{0,0-0},{0}},{{0,0-0},{0},0}

{{0,0-0},{0},{0}} returns {{0,0-0},{0}<,0>}

And so on.

{0,{0,0-0}} returns: {{0,0-0}<,{0,0-0}>}
{{{0,0-0}}} returns: {<{0->{0}<}>,{0,0-0}}
{{{{0,0-0}}}} returns: {{{<{0->{0}<}>,{0,0-0}}}}

{0-{0,0-0}} returns: <{>0,0-0<}>

So:

3{0-{0,0-0}} = {{{{0,0-0}}}} = {{{{0-{0-{0-{0-{0}},{0,0-0}}}}

...

So you were right, these don't match what I've said in previous posts. But still, {0-{0,0-0}} dominates everything that doesn't have a "0-{0,0-" somewhere, and 3{0-{0,0-0}} should go past f_φ(2,φ(3,0)+1)(3) in the FGH.

There's also these series:

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

Which dominate anything without a "0-{0,0-", "0-{0-{0,0-" "0-{0-{0-{0,0-"...

And:

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

Which has to go again:

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

And once you run out of things to do you add a 0- so in {0-0-0} you have to nest bigger values to reach {0-0-0,0}.

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

Postby Deedlit » Sat Mar 07, 2015 9:03 am UTC

Thanks for your effort, but it would really help if you could create a set of general rules for your notation. It looks like {X} means something vaguely like "reduce the least important thing and iterate at the next lower level", but that needs to be made precise.

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

Postby Vytron » Sat Mar 07, 2015 9:17 am UTC

It would help if you mentioned one string that you don't know how to reduce so I explain, because I don't know what's unclear.

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

Postby Vytron » Sat Mar 07, 2015 10:54 am UTC

Okay, let's try this:

We have:

n,a,b,c,d...

Which is n followed by a , followed by elements, each separated by a ,

Find the highest element, call it h.

Look and see if there's an element as high as h.

If it is, call it h2, define:

n,x,s

Where x will remain the same, and s is a string we will reduce.

h2 is the last element of x.

Otherwise, x doesn't exist and we only have:

n,s

Reducing s:

n,s is a string a,b,c,d... where there's some h that is the highest element.

For n,h (i.e. h is the only element):

n,0 = <(>n<),0>
n,{0} = n<,0>
n,{0,0} = n<,0>,{0}
n,{0,0,0} = n<,0>,{0},{0,0}
...

That is, any {X} where X is just a string of 0s reduces to n+1 copies of ,0 followed by {0},{0,0},{0,0,0}... up until you have n-1 0s in the last {X}

For n,a,h (i.e. h is preceded by a string where each element is less than h):

Reduce a as above, and return n<,a,h>

Examples:

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

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

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

Beyond {{0}}:

To add a ,0 at the right of X in {{0}}, we have to reach the value to where X would collapse to at the left of {X}. Let's call it Y, we return n<,Y,{X}>

For {{0}} it's {0<,0>}, so we have:

n,{{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}>

That is, nested 0s list every value from right to left until the one without any nested 0s is arrived at.

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

Now, {{0}<,0>} would be the value reached without nested 0s, so nested 0s return:

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

Other than this, n,{X} follows the same progression of n,X, so its limit is:

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

{0-0} works the same, so we have:

n,{{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}}>

And since n,{X} follows the same progression of n,X, its limit is:

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

Now, generally speaking, n,{0-X} evaluates to n,<{>0-X<}>, where X is a reduced string, if X is a string less than {0,0-0}.

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

{0,0-0} is the biggest element yet, so when you find it in n,{0-X}, you return n,<{>X<}> instead.

<{0->{0}<}> is the element that {0,0-0} would reduce to, so nested 0s return:

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}>

And since n,{X} follows the same progression of n,X, its limit is:

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

And, generally, n,{0,0-X} = n,<{0->{0,0-X}<}> where X is a reduced string, n,{Y-X} = n,<{y->{Y-x}<}> where lowercase indicates a reduces string, if Y>X, but for X<Y then n,{Y-X} = n,<{y->{y-X}<}>.

And:

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

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

Are these rules general enough, because I have no idea how else to write them.

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

Postby Deedlit » Sat Mar 07, 2015 1:20 pm UTC

Not quite general enough, but it's a good start.

Of course, the notion of "bigger" will have to be precisely defined, but usually I have a pretty good idea of what is bigger.

[quote}
And, generally, n,{0,0-X} = n,<{0->{0,0-X}<}> where X is a reduced string, n,{Y-X} = n,<{y->{Y-x}<}> where lowercase indicates a reduces string, if Y>X, but for X<Y then n,{Y-X} = n,<{y->{y-X}<}>.
[/quote]

I think you wrote that backwards - you want n,{Y-X} = n,<{y->{Y-x}<}> if X < Y, but you want n,{Y-X} = n,<{y->{y-X}<}>. if Y < X.

So for example, 3,{0->{0,0-0},0} will resolve to {{{0->{0,0-0}}}}, which will resolve to... well I guess you put {{0->{0,0-0}}} on the right, and resolve it on the left - so you get n, <<{>0,0-0<}> {0->{0,0-0}}> {{0->{0,0-0}}}, I guess.

But wait, this is the same thing as what you get with {{0->{0,0-0}},0}. Which of {{{0->{0,0-0}}}} and {{0->{0,0-0}},0} is bigger, and how does one resolve to the other?

Note that to make a general ruleset, you need to say how

{a_1 - a_2 - ... a_n}

and

{a_1, a_2, ..., a_n}

and

a_1, a_2, ...., a_n

evaluate for any notations a_1,...,a_n. So far you have only explained subsets.

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

Postby mike-l » Sat Mar 07, 2015 3:10 pm UTC

Ok, so I can give standard rules for the beginning of Vytron's notation:

<X> means repeat X n times
[] is just for grouping
-- is decrement operator

Always match the first pattern that applies. "Less than" is not entirely defined yet.

n,X,0 = <(>n,<X)> (ie a 0 on the end just iterates the function)
n,X = n,X--

[X,b,Y,a]-- = X,b,[Y,a]-- where b >= a, Y is made of up terms less than a (note he hasn't quite been explaining this the same way)

0-- = nothing
[Y,a]-- = <Y--,a,> if Y is nonempty, less than a
{0} -- = 0,<0>
{X,0}-- = <{X}--, {X}> (repetition doesn't actually change ordinal growth here but it's what he's using)

{X}-- = {X--}, X does not contain any terms a-b

0-0 -- = <{> 0 <}>

To this point everything seems to match the examples, but applying the above would do
{{0-0}}-- = { {0-0}-- } = { {0-0 --} } = <{> 0 <}> again
while he wants < <{>0<}>, {0-0} >, which is < {0,0}--,{0-0} >, ie the rule that normally would be { {0-0}, 0 }, which he instead sends to <{> 0-0 <}>. On one hand this is good because it makes use of the otherwise useless <{> 0-0 <}>, but it does require some explanation of how the rules work when you have a-b terms in brackets beyond "follows the rules above".
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Re: Your number is, in fact, not bigger!

Postby Vytron » Sat Mar 07, 2015 10:59 pm UTC

mike-l wrote:On one hand this is good because it makes use of the otherwise useless <{> 0-0 <}>, but it does require some explanation of how the rules work when you have a-b terms in brackets beyond "follows the rules above".


I'm thinking it may not be worth it, if keeping <{> 0-0 <}> useless is easier to explain and it's known to reach the SVO. So let's do it your way.

(thanks mike for these genral rules:)

We begin at n,{0-0}, where {0- is an unchanging string.

For X in n,{0-X} we have:

<X> means repeat X n times
[] is just for grouping
-- is decrement operator

Always match the first pattern that applies.

A string n,n,n < {n,n,n} < {{n,n,n}}

n,{0-X,0} = <(>n<,{0-X})>,{0-X} (ie a 0 on the end just iterates the function)

n,{0-X} = n,{0-X--}

Reduction rules:

[X,b,Y,a]-- = X,b,[Y,a]-- where b >= a, Y is made of up terms less than a

0-- = nothing
[Y,a]-- = <Y--,a,> if Y is nonempty, less than a
{0} -- = 0,<0>
{X,0}-- = <{X}--, {X}>

{X}-- = {X--}, X does not contain any terms a-b

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

So:

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

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

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

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

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

Would this be clear enough and strong enough to reach the SVO?

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

Postby mike-l » Sat Mar 07, 2015 11:37 pm UTC

It's highly unlikely that this gets anywhere near the small Veblen ordinal. At a guess I'd put the limit of this notation at phi(4,0). But the rules seem quite incomplete so it's hard to say.
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Re: Your number is, in fact, not bigger!

Postby Deedlit » Sat Mar 07, 2015 11:44 pm UTC

No, I'm afraid not. Under the new rules if {0-X} has ordinal a, then {0-X,0} has ordinal a+1. This is actually weaker than before. So we have

{0-0} = z_0
{0-0,0} = z_0 + 1
{0-{0}} = z_0 + w
{0-{{0}}} = z_0 + e_0
{0-{0-0}} = z_0 2
{0-{0-{0-0}}} = z_0 3
{0,0-0} = z_0 w
{0,0-0,0} = z_0 w + 1
{0,0-{0,0-0}} = z_0 w2
{0,0-{0,0-{0,0-0}}} = z_0 w3
{0,0,0-0} = z_0 w^2
{{0}-0} = z_0 w^w
{{0}}-0} = z_0 e_0
{{0-0}-0} = z_0^2
{{{0-0}-0}-0} = z_0^3
{0-0-0} = z_0^w = w^w^(z_0+1)
{0-0-{0-0-0}} = w^w^(z_0+1) 2
{0-0,0-0} = w^(w^(z_0+1) + 1)
{0-{0-0-0}-0} = w^(w^(z_0+1) + w^w^(z_0+1)) = w^w^w^(z_0+1)
{0,0-0-0} = e_{z_0+1}
{0,0,0-0-0} = e_{z_0+2}
{{0}-0-0} = e_{z_0+w}
{{0-0-0}-0-0} = e_{w^w^{z_0+1}}
{{{0-0-0}-0-0}-0-0} = e_e_{w^w^{z_0+1}}
{0-0-0-0} = z_1
{0,0-0-0-0} = z_2
{{0-0-0-0}-0-0-0} = z_z_1
{0-0-0-0-0} = phi(3,0)
{0-0-0-0-0-0} = phi(4,0)

So the notation still has a limit of phi(w,0).

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

Postby Vytron » Sat Mar 07, 2015 11:45 pm UTC

I'm trying to mimic your rules from this post, with -s instead of |s, and {}s instead of []s, and I quote:

mike-l wrote:Anyway, here's a number that's certainly much bigger than Vytron's. I still haven't looked much at WarDaft's, but this should fall pretty short of his claims and only be about Gamma_0, unless my analysis of Vytron's number above is wrong. In any case, this subsumes Vytron's notation.

So the main difficulty in Vytron's notation is coming up with new ways of putting terms together. I'm simply going to automate this process. Basically I'm going to enumerate types of parentheses.

A term is either 0 or [T|T,T...T] where each T is a term. The T before the | is the level of the term, and they are ordered so that the level takes precedence (if a term is a higher level, then it is higher overall).

[0| X] behaves exactly as Vytron's +s and {} do, though for simplicity I'll treat terms as functions start at 0(n) = n+1.

So
[0|X,0](n) = [0|X]^n(n), where X is any sequence of terms, with the convention that [0|X] is 0 if X is empty.

Clearly [0|0,0,...0] is just f_k.

But then you can put any of these as terms, so we have our braces, you can do things like (using the same reduction rules as Vytron)

[0| [0|0]](n) = [0|0,0,...0](n) (Corresponding to Vytron's n+{0}+ = n+0+0+...0+)
[0| 0, [0|0]](n) = [0|[0|0],[0|0],...,[0|0]](n) (Corresponding to Vytron's n+0+{0}+ = n+{0}+{0}+...+{0})

And of course you can nest within nesting, so
[0| [0| [0| ... [0|0] ... ]]] corresponds to {{...{0}...}}

Vytron then introduces concatenation as the next way of combining terms, and I cover all of this and more by just incrementing the prefix.

[ [0|0] | 0 ](n) = [0| [0| [0| ... [0|0] ... ]]]

Since Vytron's notation can't nest concatenations (this is different than concatenating with nested elements), this already surpasses his power. But I can put any term I want at all as a prefix.

So [ [0|0,0] | 0 ](9) should handily beat Vytron's number, but I can define F(n) = [...[[[0|0] | 0] | 0]... | 0] (note F gives back a term, which is a function). Then F(n)(n) should grow at Gamma_0, and I can send F(9)(9).

Allowing multiple prefixes gets to the Small Veblen Ordinal, and adding indexes to the prefixes should get to the Large Veblen. This still falls short of WarDafts claims though.


(Admittedly I've been very non-explicit about how any of the reduction works)


(emphasis mine)

So what am I doing wrong?

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

Postby Deedlit » Sun Mar 08, 2015 12:21 am UTC

Hmm, it looks like in that post mike-l never describes how he handles [X | Y,0] when X is anything other than 0, and how he handles that is key to whether his notation reaches Gamma_0 or just zeta_0.

If he uses the same rule for X as for 0, i.e. [X | Y,0] (n) = [X | Y]^n (n), then he'll only get to zeta_0, and the extension to arbitrarily many variables will only get to phi(w, 0). The key is that the Veblen notation (and the Extended Veblen notation) represent rapidly increasing functions of the rightmost variable, not static functions. So what happens when you increase the rightmost variable depends very much on all the variables that precede it.

Here's how the extended Veblen function reduces:

if a is limit, phi(X, a, O) [n] = phi(X, a[n], O) {where X is any sequence and O is a sequence of zeroes}
phi(O, a+1) [n] = phi(O, a) * n

phi(X, a+1, 0, O) [0] = 0
phi(X, a+1, 0, O) [n+1] = phi(X, a+1, phi(X, a+1, 0, O) [n], O)

phi(X, a+1, O, b+1) [0] = phi(X, a+1, O, b)+1
phi(X, a+1, O, b+1) [n+1] = phi(X, a, phi(X, a+1, O, b+1) [n], O)

if a is limit, phi(X, a, O, b+1) [n] = phi(X, a[n], phi(X, a, O, b) + 1, O)

So you can make your notation similar to that, or you can do simplify it a bit like in WarDaft's and my notations.

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

Postby mike-l » Sun Mar 08, 2015 12:49 am UTC

Yeah my post there intentionally left out many details.
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Re: Your number is, in fact, not bigger!

Postby Vytron » Sun Mar 08, 2015 1:29 am UTC

Well, okay, I'll give up now. I don't think all the time I've spent trying to find an answer that someone else already knows is worth it. Mike has gone a great length analyzing my notation and noting its problems, yet, he knows how to fix those problems but keeps them to himself because I'm supposed to "figure it out myself" or something.

He could have shown a way for my notation to reach the Large Veblen, but intentionally kept it to himself, instead choosing to sit back and watch me try and fail to reach the SVO.

Thanks Deedlit for all your job, I'm quitting because I'm feeling like a circus animal or a monkey in a cage that people go and see try and fail.

People watching know that the notation can reach:

n,{0-0} = φ(1,0)
n,{0,0-0} = φ(2,0)
n,{{0}-0} = φ(ω,0)
n,{{{0}-0}-0} = φ(φ(ω,0))
n,{0-0-0} = φ(1,0,0)

With the right reduction rules, but don't tell me what those are.

This would be understandable if they had a number on the thread that they didn't want me to reach, as, it'd make sense because by keeping this info to themselves, I'll never beat them. But mike isn't even in the competition, he enjoys knowing something I don't and bragging about it in my face. It's like I'm stuck in the mud and he gets stuck in the mud and leaves it to show it's possible, but never tells me how he did it.

Expected:

Your number doesn't reach what you want it to reach because X. However, it's possible for it to reach what you want if you did things differently.

For instance, if you did it like Y it would work.


Gotten:

Your number doesn't reach what you want it to reach because X. However, it's possible for it to reach what you want if you did things differently.

I'm intentionally leaving out the details of how you could do it for it to work.


So I don't even know what's the point of this is anymore. I'll probably stick to beating Daggoth since he's still sending values I know how to pass.

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

Postby Vytron » Sun Mar 08, 2015 1:51 am UTC

This would also have been fine:

Your number doesn't reach what you want it to reach because X, and because of Y, it's impossible for it to ever reach it.


That it's known it's possible and it's known how to do it, but it's kept secret, it's what bothers me.

So I'll just send:

3|(0[0;]0)

In WarDaft's notation.

Because if the thread isn't "let's invent a strong notation together" because help on how to do it is restricted by secrets, then trying and failing to create it is as pointless as just +1ing someone else's notation.

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

Postby Daggoth » Sun Mar 08, 2015 1:54 am UTC

I'm struggling to reach ordinals even lower than your's and i dont see you or anyone helping me out at all.

(by the way you also pass the fact that Deelit just gave you a blueprint for SVO)
Last edited by Daggoth on Sun Mar 08, 2015 2:26 am UTC, edited 1 time in total.

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

Postby mike-l » Sun Mar 08, 2015 2:01 am UTC

I didn't realize this thread was "help Vytron make a bigger number". And given that you're using the base reduction rules I gave you your accusations fall pretty flat.
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Re: Your number is, in fact, not bigger!

Postby Deedlit » Sun Mar 08, 2015 2:52 am UTC

Woah. Vytron, I intentionally didn't give you precise rules for how to make your notation reach the SVO, not because I didn't want you to reach it, but because I figured you wanted to make your own notation, and giving you the exact rules might get in the way of that. I would certainly have given you rules if you had asked.

For example, using the rules for the extended Veblen function as a guide, we can do this:

X stands for an arbitrary sequence x1-x2-...-xn
O stands for an arbitrary sequence of zeroes 0-0-...-0

case 1: the rightmost nonzero variable doesn't end in ,0

then reduce it: {X-a-O}-- = {X- a-- -O}

case 2: all the variables are zero

{0-O}-- = <{><-O}>

case 3: all the variables are zero except the rightmost variable

{0-O-a,0} = <{>{0-O-a}><-O}>

For the remaining cases, let b be the rightmost variable and a be the rightmost nonzero variable to the left of a.

case 4: b = 0

{O1 - a,0 - 0 - O2}-- = <{O1 - a -> 0 <-O2}>

case 5: a and b both end in ,0

{O1 - a,0 - O2 - b,0} = <{O1 - a - >{{O1 - a,0 - O2 - b}} <- O2}>

case 6: b ends in ,0, a does not

{O1 - a - O2 - b,0} = {O1 - a-- - {{O1 - a - O2 - b}} - O2}

That should do it.

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

Postby Deedlit » Sun Mar 08, 2015 3:13 am UTC

Oh, and Daggoth, sorry that I haven't commented on your notations, but I haven't had the energy to try and decipher them in addition to Vytron's notation.

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

Postby WarDaft » Sun Mar 08, 2015 3:26 am UTC

Vytron wrote:People watching know that the notation can reach:

n,{0-0} = φ(1,0)
n,{0,0-0} = φ(2,0)
n,{{0}-0} = φ(ω,0)
n,{{{0}-0}-0} = φ(φ(ω,0))
n,{0-0-0} = φ(1,0,0)


The way you do that is by basically re-inventing the Veblen notation in your own words. There's not really any way around that.
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Re: Your number is, in fact, not bigger!

Postby Vytron » Sun Mar 08, 2015 3:49 am UTC

Thanks Deedlit! That's the kind of answer I was hoping for when I asked "what was I doing wrong"! This notation is already not mine, back when mike generalized the rules of prefixes I said it was the "mike-l-Vytron" notation, and I went with it because it worked. When I said "So let's do it your way" I was giving up on extending my notation the way I wanted because my way wasn't reaching the SVO.

I started another attempt but after hitting preview I saw mike already knew the reduction rules but deliberately left them out, so deleted it and quit instead.

I don't think I'd have ever arrived to quite the same rules you have posted, so I appreciate the help.

With your reduction rules this would be the "Deedlit-mike-l-Vytron" notation.

Let's see now how these evaluate:

3,{{0-0}-{0,0}-0,0-0-0,0,0}

The rightmost variable is b, and the third variable (the firs one not 0 at the left of b) is a.

a=0,0
b=0,0,0

Case 5 holds (a and b both end in ,0)

Reduction rules:

{O1 - a,0 - O2 - b,0} = <{O1 - a - >{{O1 - a,0 - O2 - b}} <- O2}>

By replacement:

O1={0-0}-{0,0}
O2=0

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

3,{{0-0}-{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}

And:

3,{{0-0}-{0,0},0-{0,0}-0-0-0,0,0}

The rightmost variable is b, and the third variable (the firs one not 0 at the left of b) is a.

a={0,0}
b=0,0,0

Case 6 holds (b ends in ,0 - a doesn't)

Reduction rules:

{O1 - a - O2 - b,0} = {O1 - a-- - {{O1 - a - O2 - b}} - O2}

O1={0-0}-{0,0},0
O2=0-0

By replacement:

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

3,{{0-0}-{0,0},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}

If that's it then, yay! Something that works at last! :)

Now, I'll be looking forward to an analysis of Daggoth's notation. I DO have taken a look at it, and think replacing ¿?s for []s and ¡!s for ()s does help with its deciphering, just look at these equivalent values:

¿0¡0¡1!0¡1!0!0?10

[0(0(1)0(1)0)0]10

I find it much easier to see what's going on in the latter for some reason.

I'm afraid he doesn't reach e_1, though. That's because he has:

[0(0(0)0(0)0(1)0)0]10 > fε0^ε0^ω^ω(10)

However, by his definitions, I believe what should be f{ε0^ε0}^ω^ω(10) not fε0^{ε0^{ω^ω}}(10) because he has defined the progression as x^ω^ω, x^ω^ω^ω, x^ω^ω^ω^ω, its limit would be f{ε0^ε0}^ε0(10), not f{ε0^ε0^ε0}(10)

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

Postby Daggoth » Sun Mar 08, 2015 4:34 am UTC

Adding ¡0!'s to the right side gives the x^ω, x^ω^ω, x^ω^ω^ω progression. +1 on the left side of ¡1! is the limit of that, x^ε0,

So piling up +1's at the left of ¡1! is (...(ε0^ε0)^ε0)...^ε0. Since (a^b)^c = a^b*c, these )^ε0's "collapse" down like this ε0^(ε0*ε0...*ε0) a.k.a ε0^ε0^ω

a 0¡0!0 on the left is the limit of these +1's, ε0^ε0^ω. SO far so good

b¡0!a here
a: f#^ε0 limits at b+1 #^ε0^ω
b: f#^ε0^ω

b+1: ((x^ε0^ω)^ε0^ω) = x^ε0^ω^2
b+2: ((x^ε0^ω^2)^ε0^ω) = x^ε0^ω^3
limit 0¡0!0¡0!0 = x^ε0^ω^ω
c¡0!b¡0!a
a: f#^ε0 limits at b+1 #^ε0^ω
b: f#^ε0^ω
c: f#^ε0^ω^ω


Right side <¡0!0> progression:
x^ω,x^ω^ω,x^ω^ω^ω,
Left side <¡0!0> progression:
#^ε0^ω,#^ε0^ω^ω,#^ε0^ω^ω^ω
Left side ¡0! progression limits at 0¡1¡0¡1!0 ε0^ε0^ε0

(unless i'm horribly wrong, you did make me double-check so thanks for that anyway)

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

Postby Deedlit » Sun Mar 08, 2015 4:35 am UTC

I can't say about Daggoth's notation, because I don't see a definition, it must be pretty far back in the thread.

Okay, now seems like a good time to challenge WarDaft's leading notation, which I believe is currently at the level psi(Ω^Ω^Ω^2).

Okay, so this time I'll avoid the use of numbers in the main expressions, although they could certainly be inserted if one considers that prettier. Notations will consist of a number followed by balanced pairs of parentheses and brackets, so

83|([(()[()()])([])]()())

is a valid notation. One rule: brackets are never on the outside, they are always contained in parentheses or other brackets.

The rules are:

n| = n
n|X = n+1|X--

X()-- = X
X(Y())Z-- = X<(Y)>Z, where Z consists of only ) and ]
X[Y()]Z-- = X<[Y]>Z, where Z consists of only ) and ]
X(Y[]W)Z-- = X<(Y>()<W)>Z, where Z consists of only ) and ], and W consists of only ]

and that's it!

Describing it with words: To reduce an expression, find the rightmost () or []. If it is a (), remove it and replicate the immediately surrounding pair of parentheses/brackets, along with their contents, n times; if there is no surrounding pair of parentheses/brackets, just remove it. If it is a [], find the nearest enclosing (), replace the [] with the entire enclosing () n times, then finally replace the [] with an empty ().

Examples:

3|([]([][])(([])()) = 4|([]([][])<(([]))> = 4|([]([][])(([]))(([]))(([]))

3|([]())(()([])) = 4| ([]())<(()(>()<))> = 4| ([]())(()((()((()(()))))))

Analysis:

The notation with parentheses alone is simply my earlier notation going up to epsilon_0. More generally, (x1 x2 ... xn) corresponds to the function psi(x1 + x2 + ... + xn), where the psi function is like the usual one for the Bachmann-Howard ordinal, except it starts with psi(a) = w^a, psi(Ωa) = epsilon_a, psi(Ω^2 a) = zeta_a, etc. (psi(Ω^Ω a) is still Gamma_a.) [x1 x2 ... xn] corresponds to the function Ω^(1 + x1 + x2 + ... + xn).

So:

() = 1
()() = 2
(()) = w
(())(()) = w2
(()()) = w^2
((())) = w^w
(((()))) = w^w^w
([]) = epsilon_0
([]()) = w^(epsilon_0 + 1)
([][]) = epsilon_1
(([][][]())) = w^w^(epsilon_2 + 1)
([()]) = zeta_0
([()()]) = phi(3,0)
([(())]) = phi(w,0)
([((()))]) = phi(w^w,0)
([([])]) = phi(epsilon_0, 0)
([([()])]) = phi(zeta_0, 0)
([[]]) = Gamma_0
([[][]]) = phi(2,0,0)
([[()]]) = phi(1,0,0,0)
([[()()]]) = phi(1,0,0,0,0)
([[(())]]) = SVO = psi(Ω^Ω^w)
([[[]]]) = LVO = psi(Ω^Ω^Ω)
([[[[]]]]) = psi(Ω^Ω^Ω^Ω)

So, I guess I will submit the number 9([[[[]]]]), which should be the current champion.
Last edited by Deedlit on Sun Mar 08, 2015 4:40 am UTC, edited 1 time in total.

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

Postby Deedlit » Sun Mar 08, 2015 4:39 am UTC

Daggoth wrote:
b+1: ((x^ε0^ω)^ε0^ω) = x^ε0^ω^2
b+2: ((x^ε0^ω^2)^ε0^ω) = x^ε0^ω^3
limit 0¡0!0¡0!0 = x^ε0^ω^ω
c¡0!b¡0!a
a: f#^ε0 limits at b+1 #^ε0^ω
b: f#^ε0^ω
c: f#^ε0^ω^ω


Actually, ((x^ε0^ω)^ε0^ω) = (x^(ε0^ω)*(ε0^ω)) = (x^(ε0^(ω+ω))) = x^ε0^(ω2), not x^ε0^ω^2.

so

b+2: ((x^ε0^(ω2))^ε0^ω) = x^ε0^(ω3)
limit 0¡0!0¡0!0 = x^ε0^ω^2

instead

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

Postby Vytron » Sun Mar 08, 2015 4:49 am UTC

Question:

Does it continue:

([[[[]]]])=ψ(Ω^Ω^Ω^Ω)
([[[[[]]]]])=ψ(Ω^Ω^Ω^Ω^Ω)
([[[[[[]]]]]])=ψ(Ω^Ω^Ω^Ω^Ω^Ω)

Making the limit ψ(ε_{Ω+1})?

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

Postby Deedlit » Sun Mar 08, 2015 4:57 am UTC

Vytron wrote:Question:

Does it continue:

([[[[]]]])=ψ(Ω^Ω^Ω^Ω)
([[[[[]]]]])=ψ(Ω^Ω^Ω^Ω^Ω)
([[[[[[]]]]]])=ψ(Ω^Ω^Ω^Ω^Ω^Ω)

Making the limit ψ(ε_{Ω+1})?


Yup!

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

Postby Vytron » Sun Mar 08, 2015 6:01 am UTC

That kind of boggles the mind. I think I'd have to add a new extension for each ^Ω so I'm going to do some research to see how I can take a shortcut to that growth...

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

Postby Vytron » Sun Mar 08, 2015 11:01 am UTC

Okay, no shortcuts here because from what I read shortcuts mean getting rid of everything I have so far and "going for something stronger" or something.

So I'll make:

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

So that it reaches the SVO.

Now {0[0]0} can be used in any place of n,{0<-0>} for the same rules.

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]0,0}>}

And in general:

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

So there's:

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

And:

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

X stands for an arbitrary sequence x1-x2-...-xn
O stands for an arbitrary sequence of zeroes 0-0-...-0

Case 1: The rightmost nonzero variable after {0[0] doesn't end in ,0

n,{0[0]X-a-O}-- = n,{0[0]X- a-- -O}

Case 2: all the variables after {0[0] are zero

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

Case 3: all the variables after {0[0] are zero except the rightmost variable:

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

For the remaining cases, let b be the rightmost variable and a be the rightmost nonzero variable to the left of a and to the right of 0[0].

Case 4: b = 0

n,{0[0]O1 - a,0 - 0 - O2}-- = n,<{0[0]O1 - a -> 0 <-O2}>

Case 5: a and b both end in ,0

n,{0[0]O1 - a,0 - O2 - b,0} = n,<{0[0]O1 - a - >{{0[0]O1 - a,0 - O2 - b}} <- O2}>

case 6: b ends in ,0, a does not

n,{0[0]O1 - a - O2 - b,0} = n,{0[0]O1 - a-- - {{0[0]O1 - a - O2 - b}} - O2}

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

n,{0,0[0]0} uses the same rules, but duplicates {0[0]{0,0[0]s on the inner nestings instead of {{0[0]s

n,{0,0,0[0]0} uses the same rules, but duplicates {0,0[0]{0,0,0[0]s on the inner nestings instead of {0{0,0[0]s

And in general:

n,{X[0]0} uses the same rules, but duplicates {X--[0]{X[0]s on the inner nestings

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

Now, let's define:

X stands for an arbitrary sequence x1[0]x2[0]...[0]xn
O stands for an arbitrary sequence of zeroes 0[0]0[0]...[0]0

Case 1: The rightmost nonzero variable doesn't end in ,0

n,{X[0]a[0]O}-- = n,{X [0] a-- [0] O<-0>}

Case 2: all the variables are zero

{0[0]O}-- = <{><[0]O}>

Case 3: all the variables are zero except the rightmost variable

{O[0]a,0} = <{>{0[0]O[0]a}><-O}>

For the remaining cases, let b be the rightmost variable and a be the rightmost nonzero variable to the left of a.

Case 4: b = 0

{O1 [0] a,0 [0] 0 [0] O2}-- = <{O1 [0] a [0]> 0 <[0]O2<-0>}>

case 5: a and b both end in ,0

{O1 [0] a,0 [0] O2 [0] b,0} = <{O1 [0] a - >{{O1 [0] a,0 [0] O2 [0] b}} <[0] O2}>

case 6: b ends in ,0, a does not

{O1 [0] a [0] O2 [0] b,0} = {O1 [0] a-- [0] {{O1 [0] a [0] O2 [0] b}} [0] O2}

So, the main difference here is the adding of <-0> at the right of stuff.

3,{{0-0}[0]{0,0},0[0]{0,0}[0]0[0]0[0]0}

The rightmost nonzero variable doesn't end in ,0, this is case 1:

n,{X[0]a[0]O}-- = n,{X [0] a-- [0] O<-0>}

By replacement:
X={0-0}[0]{0,0},0
a={0,0}
O=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},0[0]0<,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} = 3,{{0-0}[0]{0,0},0[0]0,0,0,0{0}[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}

And:

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

Now, I wonder if this is strong or useless.

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

Postby WarDaft » Sun Mar 08, 2015 5:48 pm UTC

Very well then,

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)>
(0[:]1)-- = (<0[><:]0>)
(a,0[:]b)-- =(<0[>((a--,0[:]b))<:]b-->)

Then 2|(2,0[:]1) to be juuust ahead of where Deedlit can go without changing things. Specifically, ψ(εΩ+1*3)
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Re: Your number is, in fact, not bigger!

Postby Daggoth » Sun Mar 08, 2015 8:09 pm UTC

Heres GFDQM With hopefully the wording and examples made 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 separators. What they do is basically separate two numbers.
@ Indicates a cascade. for an example see step 5

Base cases:
Step 1: ¿0?n = n↑nn // ω
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
@ = @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!, both to the right and left
example: ¿1¡2!0?3
Spoiler:
<¿0¡2!3¡0!3¡0!3¡0!3¡1!3¡0!3¡0!3¡0!3¡1!3¡0!3¡0!3¡0!3¡1!3¡0!3¡0!3¡0!3?>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

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

Postby Vytron » Sun Mar 08, 2015 11:14 pm UTC

@WarDaft: Do you reach the SVO at (0[:]0,X) = (<0,>X)?

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

Postby Deedlit » Sun Mar 08, 2015 11:21 pm UTC

WarDaft wrote:Very well then,

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)>
(0[:]1)-- = (<0[><:]0>)
(a,0[:]b)-- =(<0[>((a--,0[:]b))<:]b-->)

Then 2|(2,0[:]1) to be juuust ahead of where Deedlit can go without changing things. Specifically, ψ(εΩ+1*3)


These rules seem incomplete. The (X [Y:] Z) rules are only defined for when X and Y are at most two variables. What if X has more than two variables? Some of the time you can use eariler rules, but how about (a,X,0,b,Y) when b contains square brackets?

Also, if Y never has more than two variables, you can't go to notations like (0[0[:]0:]0).

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

Postby Deedlit » Mon Mar 09, 2015 6:49 am UTC

Vytron wrote:Okay, no shortcuts here because from what I read shortcuts mean getting rid of everything I have so far and "going for something stronger" or something.

So I'll make:

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

So that it reaches the SVO.

Now {0[0]0} can be used in any place of n,{0<-0>} for the same rules.

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]0,0}>}

And in general:

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


This looks good.

So there's:

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


This doesn't look so good. Using the above rules, this should be

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

And:

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


So we're making [0] a higher precedence operator than -? That's fine.

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


So this would be at phi(2@w).

n,{X[0]0} uses the same rules, but duplicates {X--[0]{X[0]s on the inner nestings


if X has ordinal a, {X[0]0} would have ordinal phi(a@w)

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


n,{0[0]0[0]0} has ordinal phi(1@w+1)

{0[0]0[0]0[0]0} has ordinal phi(1@w+2)

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


{0-0[0]0} has ordinal phi(1@w2)

Now, I wonder if this is strong or useless.


Not a bad start. My first thought was to have 0-0[0]0 do what your 0[0]0[0]0 currently does, so that that would be phi(1@w+1), 0[0]0[0]0 would have ordinal phi(1@w2), and the notation would go up to phi(1@w^2). But you still have more dashes left, so maybe you can make something equal or stronger. Please continue.

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

Postby Vytron » Mon Mar 09, 2015 4:01 pm UTC

Thanks for your analysis Deedlit. Yesterday's night I had a revelation. If what is getting me to the SVO is stronger rules, can't I just "cheat" by using these rules earlier? In your analysis you point out I "run out of dashes earlier" doing it like this, so there's a better ordering of dashes, implying saving dashes is important.

(Note, the plan was to have something that follows my original progression mixing the dashes and [0]s, for instance:

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

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

So the -s work like 0s in the base progression and [0]s work like {0}s in the base progression.

Then once I have:

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

That means the dash is removed and we go to:

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

And:

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

But I want to explore the idea below first)

For my implementation of the rules earlier I'd do it this way:

Keep the rules up to n,{0-0} = n,<{>0<}>

Now, we define:

n,{0-0},{{0}} = <(>n<,{0-0})>,{0-0}

And we apply the rule changes:

Rules for Y in n,{0-0},{Y}

X stands for an arbitrary sequence {x1},{x2},...,{xn}
O stands for an arbitrary sequence of enclosed 0s {0},{0},...,{0}

Case 1: the rightmost non{0} variable doesn't end in ,0}

then reduce it: {X,{a},O}-- = {X, {a--} ,O}

Example:

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

{{0}} doesn't end in ,0}

X={0},{0,0},
a={0}
O=,{0},{0}

By replacement:

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

case 2: all the variables are {0}

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

case 3: all the variables are {0} except the rightmost variable

n,{0-0},{{0},O,{a,0}} = n,{0-0}<{{0-0},{>{0-0},{{0}-O-a}><-O}>

For the remaining cases, let b be the rightmost variable and a be the rightmost non{0} variable to the left of a.

case 4: b = {0}

n,{0-0},{O1,{a,0},{0},O2}-- = n,{0-0},{<{0-0},{O1,{a},> {0} <,O2}>

case 5: a and b both end in ,0}

n,{0-0},{O1,{a,0},O2,{b,0}} = n,{0-0},<{{0-0},{O1,{a},>{{0-0},{{0-0},{O1,{a,0},O2,{b}}} <,O2}>

case 6: b ends in ,0} a does not

n,{0-0},{O1,{a},O2,{b,0}} = n,{0-0},{O1,{a--},{{0-0},{{0-0},{O1,{a},O2,{b}}},O2}

Now, suppose I make:

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

Would that reach the SVO?

The notation would continue adding stronger and stronger rules:

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

But when there's ,{0-0},{0-0}, the rules duplicate {0-0},{0-0}s instead of {0-0}s

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

And of course we'd have:

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

Let's see if this is consistent with the classic:

3,{0-0},{{{0-0}},{{0,0}},{0,0},{0},{0,0,0}}

The rightmost variable is b, and the third variable (the firs one not {0} at the left of b) is a.

Case 5 holds (a and b both end in ,0})

a=0,0
b=0,0,0

Reduction rules:

n,{0-0},{O1,{a,0},O2,{b,0}} = n,{0-0},<{0-0},{O1,{a},>{{0-0},{{0-0},{O1,{a,0},O2,{b}}} <,O2}>

By replacement:

O1={{0-0}},{{0,0}}
O2={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}},{{0,0}},{0,0},{0},{0,0,0}}} <,{0}}>

3,{0-0},{{{0-0}},{{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}}

And:

3,{0-0},{{{0,0}},{{0,0},0},{{0-0}},{0},{0},{0,0,0}}

The rightmost variable is b, and the third variable (the firs one not {0} at the left of b) is a.

a={0-0}
b=0,0,0

Case 6 holds (b ends in ,0 - a doesn't)

n,{0-0},{O1,{a},O2,{b,0}} = n,{0-0},{O1,{a--},{{0-0},{{0-0},{O1,{a},O2,{b}}},O2}

By replacement:

O1={{0,0}},{{0,0},0}
O2={0},{0}

3,{0-0},{{{0,0}},{{0,0},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}}


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