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

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

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

Well, there's just one issue, and that is you want to insert notations into other notations without creating ambiguity.

For example, I want to be able to have

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

with A = {0-0}, {{0-0},{{0,0}}}, but of course we want

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

to be something else entirely.

The simplest solution I see is to enclose an extra pair of brackets around expressions with {0-0} in them, so the above would be

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

Under these rules, {{0-0},{0-0}} would be the SVO, {{0-0},{0-0},{0-0}} would be phi(2@w), and {0,{0-0}} would be phi(w@w).

For example, I want to be able to have

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

with A = {0-0}, {{0-0},{{0,0}}}, but of course we want

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

to be something else entirely.

The simplest solution I see is to enclose an extra pair of brackets around expressions with {0-0} in them, so the above would be

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

Under these rules, {{0-0},{0-0}} would be the SVO, {{0-0},{0-0},{0-0}} would be phi(2@w), and {0,{0-0}} would be phi(w@w).

- 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!

Nice, thanks, so the revised rules would be:

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}-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}}

So now:

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)

Reduction rules:

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}}}

Now, I wonder if this just gets me into a hole, because I'm nesting with weaker rules here.

n,0,{0-0},{Y} follows the rules of above, but nesting {{0,{0-0},{Y}}} instead of {{{0-0},{Y}}}

n,0,{0-0},{0-0},{Y} follows the rules of above, but nesting {{0,{0-0},{0-0},{Y}}} instead of {{0,{0-0},{Y}}}

n,0,{0-0},{0-0},{0-0},{Y} follows the rules of above, but nesting {{0,{0-0},{0-0},{0-0},{Y}}} instead of {{0,{0-0},{0-0},{Y}}}

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

phi(w@w2)

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

phi(w@w3)

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

phi(w@w^2)

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

phi(w@w^3)

And in general:

n,X,{0-0} follow the original progression, so:

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

Would be at phi(w@e_0)

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}-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}}

So now:

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)

Reduction rules:

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}}}

Now, I wonder if this just gets me into a hole, because I'm nesting with weaker rules here.

n,0,{0-0},{Y} follows the rules of above, but nesting {{0,{0-0},{Y}}} instead of {{{0-0},{Y}}}

n,0,{0-0},{0-0},{Y} follows the rules of above, but nesting {{0,{0-0},{0-0},{Y}}} instead of {{0,{0-0},{Y}}}

n,0,{0-0},{0-0},{0-0},{Y} follows the rules of above, but nesting {{0,{0-0},{0-0},{0-0},{Y}}} instead of {{0,{0-0},{0-0},{Y}}}

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

phi(w@w2)

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

phi(w@w3)

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

phi(w@w^2)

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

phi(w@w^3)

And in general:

n,X,{0-0} follow the original progression, so:

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

Would be at phi(w@e_0)

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

Vytron wrote:Now, I wonder if this just gets me into a hole, because I'm nesting with weaker rules here.

n,0,{0-0},{Y} follows the rules of above, but nesting {{0,{0-0},{Y}}} instead of {{{0-0},{Y}}}

n,0,{0-0},{0-0},{Y} follows the rules of above, but nesting {{0,{0-0},{0-0},{Y}}} instead of {{0,{0-0},{Y}}}

n,0,{0-0},{0-0},{0-0},{Y} follows the rules of above, but nesting {{0,{0-0},{0-0},{0-0},{Y}}} instead of {{0,{0-0},{0-0},{Y}}}

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

phi(w@w2)

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

phi(w@w3)

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

phi(w@w^2)

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

phi(w@w^3)

And in general:

n,X,{0-0} follow the original progression, so:

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

Would be at phi(w@e_0)

actually, I think

0,0,{0-0} is at phi(w^2@w)

0,0,0{0-0} is at phi(w^3@w)

{0},{0-0} is at phi(w^w@w)

{0-0,0} is at phi(e_0@w)

so we stay below phi(1@w+1)

A better method seems to be having

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

and

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

that would allow {X},{0-0} to be phi(a@w), and then

0,0,{0-0} = <{>0<,{0-0}}> = phi(1@w+1)

the rules for more variables would mirror the old rules without the {0-0}, and eventually we would reach

{0-0},{0-0} = <0,>{0-0} = phi(1@2w)

{0-0},{0-0}[0-0} = <0,>{0-0}{0-0} = phi(1@3w)

and the limit would be phi(1@w^2)

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

Okay, we're waiting for WarDaft to respond, but in the meantime I figured that I would increase my number again.

We will define entries to be strings of the form (x1 x2 ... xn m), where m is a nonnegative integer and the xi are entries (Call m the level of the entry). 0-entries are entries with level 0. Notations will be strings of the form i | x1 x2 ... xn, where i is a positive integer and the xj are 0-entries. (Call i the base.)

To resolve a notation, find the rightmost string of the form (m).

If m = 0, use our old () rule: If (0) is on the outside, just remove it. Otherwise, remove it, and copy the immediately surrounding entry i times, where i is the base.

If m > 0, find the closest surrounding entry with lower level than m, and nest that surrounding entry i times in (m); that is, if the notation is A(B(m)C n)D where (B(m)C n) is the closest surrounding entry of lower level, we replace it with A(<B(>m-1<)C n>)D.

Or, describing it with string rules:

n| = n

n|X = n+1|X--

X(0) -- = X

A(B(0) m)D -- = A<(B m)>D, where D does not contain any )'s

A(B(m)C n)D -- = A(<B(>m-1<)C n>)D, where n < m, C and D do not contain any )'s and C does not contain any x) with x < m.

I hope that makes sense.

The notation with just (X 0)'s correspond to the epsilon_0 notation, and notations with just (X 0)'s and (X 1)'s correspond to the Bachmann-Howard notation. In the general notation, (x1 x2 ... xn 0) represents the ordinal psi_0 (x1 + x2 + ... + xn), where again psi_0 is modified to start from psi_0(a) = w^a, psi_0(Omega a) = epsilon_a, etc. More generally, for m > 0 (x1 x2 ... xn m) represents the ordinal psi_m (x1 + x2 + ... + xn). So the notation goes to psi_0 (Omega_w).

So I'll submit the number 9 | ((9)9)0).

We will define entries to be strings of the form (x1 x2 ... xn m), where m is a nonnegative integer and the xi are entries (Call m the level of the entry). 0-entries are entries with level 0. Notations will be strings of the form i | x1 x2 ... xn, where i is a positive integer and the xj are 0-entries. (Call i the base.)

To resolve a notation, find the rightmost string of the form (m).

If m = 0, use our old () rule: If (0) is on the outside, just remove it. Otherwise, remove it, and copy the immediately surrounding entry i times, where i is the base.

If m > 0, find the closest surrounding entry with lower level than m, and nest that surrounding entry i times in (m); that is, if the notation is A(B(m)C n)D where (B(m)C n) is the closest surrounding entry of lower level, we replace it with A(<B(>m-1<)C n>)D.

Or, describing it with string rules:

n| = n

n|X = n+1|X--

X(0) -- = X

A(B(0) m)D -- = A<(B m)>D, where D does not contain any )'s

A(B(m)C n)D -- = A(<B(>m-1<)C n>)D, where n < m, C and D do not contain any )'s and C does not contain any x) with x < m.

I hope that makes sense.

The notation with just (X 0)'s correspond to the epsilon_0 notation, and notations with just (X 0)'s and (X 1)'s correspond to the Bachmann-Howard notation. In the general notation, (x1 x2 ... xn 0) represents the ordinal psi_0 (x1 + x2 + ... + xn), where again psi_0 is modified to start from psi_0(a) = w^a, psi_0(Omega a) = epsilon_a, etc. More generally, for m > 0 (x1 x2 ... xn m) represents the ordinal psi_m (x1 + x2 + ... + xn). So the notation goes to psi_0 (Omega_w).

So I'll submit the number 9 | ((9)9)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!

Deedlit wrote:A better method seems to be having

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

and

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

that would allow {X},{0-0} to be phi(a@w), and then

0,0,{0-0} = <{>0<,{0-0}}> = phi(1@w+1)

the rules for more variables would mirror the old rules without the {0-0}, and eventually we would reach

{0-0},{0-0} = <0,>{0-0} = phi(1@2w)

{0-0},{0-0}[0-0} = <0,>{0-0}{0-0} = phi(1@3w)

and the limit would be phi(1@w^2)

I see thanks.

So if I allow:

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

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

Can I reach

{0-a} = phi(1@a)

And have:

{0-{{0-0},{0-0}}} = phi(1@phi(1@w))

{0-{0-{{0-0},{0-0}}}} = phi(1@phi(1@phi(1@w)))

{0-{0-{0-{{0-0},{0-0}}}}} = phi(1@phi(1@phi(1@phi(1@w))))

?

Deedlit wrote:I figured that I would increase my number again.

Okay, it seems that I have to see it with symbols, it's:

ψ_0(Ω_ω)(9)?

How does that compare with the ordinal collapse shown in:

http://quibb.blogspot.mx/2012/03/infini ... tions.html

?

Ω_3 seems like a huge jump already and you're claiming access to Ω_ω? o_O

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

One correction to my notation: The line

A(B(m)C n)D -- = A(<B(>m-1<)C n>)D, where n < m, C and D do not contain any )'s and C does not contain any x) with x < m.

should read

A(B(m)C n)D -- = A(<B(><m-1)C> n)D, where n < m, C and D do not contain any )'s and C does not contain any x) with x < m.

Yup! So you can reach the LVO.

It should be ψ_0(ψ_9(Ω_9)).

It's equivalent.

Yep, it's pretty dizzying, considering the fact that the first epsilon number past Ω_1 allows you to reach the Bachmann-Howard ordinal, which is so much greater than epsilon_0, you can imagine how much each new level boosts the previous one. But I do reach ψ_0(Ω_ω), as the notation is designed to mirror the way the normal notation up to ψ_0(Ω_ω) works. The main difference is that the above notation doesn't have symbols for Ω_n; however, it does have versions of the ψ_n functions, and ψ_n(0) = Ω_n for n > 0, so we're just fine.

A(B(m)C n)D -- = A(<B(>m-1<)C n>)D, where n < m, C and D do not contain any )'s and C does not contain any x) with x < m.

should read

A(B(m)C n)D -- = A(<B(><m-1)C> n)D, where n < m, C and D do not contain any )'s and C does not contain any x) with x < m.

Vytron wrote:I see thanks.

So if I allow:

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

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

Can I reach

{0-a} = phi(1@a)

And have:

{0-{{0-0},{0-0}}} = phi(1@phi(1@w))

{0-{0-{{0-0},{0-0}}}} = phi(1@phi(1@phi(1@w)))

{0-{0-{0-{{0-0},{0-0}}}}} = phi(1@phi(1@phi(1@phi(1@w))))

?

Yup! So you can reach the LVO.

Deedlit wrote:I figured that I would increase my number again.

Okay, it seems that I have to see it with symbols, it's:

ψ_0(Ω_ω)(9)?

It should be ψ_0(ψ_9(Ω_9)).

How does that compare with the ordinal collapse shown in:

http://quibb.blogspot.mx/2012/03/infini ... tions.html

It's equivalent.

Ω_3 seems like a huge jump already and you're claiming access to Ω_ω? o_O

Yep, it's pretty dizzying, considering the fact that the first epsilon number past Ω_1 allows you to reach the Bachmann-Howard ordinal, which is so much greater than epsilon_0, you can imagine how much each new level boosts the previous one. But I do reach ψ_0(Ω_ω), as the notation is designed to mirror the way the normal notation up to ψ_0(Ω_ω) works. The main difference is that the above notation doesn't have symbols for Ω_n; however, it does have versions of the ψ_n functions, and ψ_n(0) = Ω_n for n > 0, so we're just fine.

- 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!

Alright, so I have:

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

For ψ(Ω^Ω^Ω)(n)

Now, I wonder if from here my notation dumbs down, since I'm just repeating the same processes instead of applying stronger processes, I wonder if my notation can get off or I get stuck adding Ωs.

We'd have:

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

And:

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

Following the same known rules (nesting{{{0,0-0},{X}}} inside)

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

So:

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

Case 2: All variables are {0}:

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

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

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

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

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

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

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

Now, let's see if I can generalize rules here.

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

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

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

...

Okay, these generalizations don't match with with what I said earlier.

I stated:

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

But by applying these rules we'd have instead:

case 5: a and b both end in ,0

Reduction rules:

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

By replacement O1 and O2 are empty, so we have:

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

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

...which should be stronger.

And now I'd have:

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

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

n,{0[0]0},{X} follows the rules

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

{X,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}>

Have:

{0[0]{{0[0]0},{0[0]0}}}

{0[0]{0[0]{{0[0]0},{0[0]0}}}}

{0[0]{0[0]{0[0]{{0[0]0},{0[0]0}}}}}

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

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

Have:

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

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

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

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

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

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

For ψ(Ω^Ω^Ω)(n)

Now, I wonder if from here my notation dumbs down, since I'm just repeating the same processes instead of applying stronger processes, I wonder if my notation can get off or I get stuck adding Ωs.

We'd have:

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

And:

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

Following the same known rules (nesting{{{0,0-0},{X}}} inside)

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

So:

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

Case 2: All variables are {0}:

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

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

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

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

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

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

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

Now, let's see if I can generalize rules here.

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

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

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

...

Okay, these generalizations don't match with with what I said earlier.

I stated:

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

But by applying these rules we'd have instead:

case 5: a and b both end in ,0

Reduction rules:

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

By replacement O1 and O2 are empty, so we have:

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

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

...which should be stronger.

And now I'd have:

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

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

n,{0[0]0},{X} follows the rules

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

{X,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}>

Have:

{0[0]{{0[0]0},{0[0]0}}}

{0[0]{0[0]{{0[0]0},{0[0]0}}}}

{0[0]{0[0]{0[0]{{0[0]0},{0[0]0}}}}}

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

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

Have:

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

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

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

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

{0,0[0]0}={0[0]0<-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!

Deedlit wrote:it's pretty dizzying

Indeed! This is like if you were at ψ(Ω^Ω^ω) and I was still catching up to e_0 all over again!

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

I'd been trying to re-find that link for AGES, thanks vyt

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

Um guys, i have an interesting quote by vytron which might explain his erratic behavior:

A killer of games is a person that posts on a thread, and then nobody posts after them, getting the game killed.

In this game, you score by killing games, intentionally or accidentally, if you were the last poster on a game and it has been dead for a while you score 1 point.

Only threads on the second page of the forum count, for the forum default (for games on the first page are deemed alive, and threads on third page and older are deemed too old to count.)

You may only post on this thread to report your points, or new points scored (when threads where you posted last go into the second page)

You may only double post (and triple post, etc.) on this thread if this thread has gone into the second page (to report that you scored in this thread by killing it.)

Games I have killed:

Something Vs. Something - Event 2!

20 Questions

Proof that Aarex is a gadfly

My number is bigger! (Aarex's version)

Where In The World Showdown!: Aarex Vs. Vytron

Your number is, in fact, not bigger!

Where In The World - V vs. TP vs. D - Time Panda

- 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!

Huh? That's just a meta-game where one gains points by being the last poster on the thread when the thread goes to page 2, it's not like one intentionally makes a post with the intention that it'll be the last post on that thread, ever (if anything, not posting on the thread would make it easier for it to be "killed" as it'd not be bumped to the top.)

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

Ignore this post, sandboxing

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

Fixed a weakness where i don't have a functional grouping indicator so i can't harness the power of nesting due to ambiguity in the notations

// indicates commentary

// the following symbols are only used to explain how the notation works and don't actually appear in the notation or its reductions

<> 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)

¡#!,¡%!, ¡@! are separators. What they do is basically separate two numbers.

@ Indicates a cascade. for an example see step 5

// The following are the actual working symbols

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

¡! indicates a separator

{} indicates grouping

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

@ = @

@

...@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

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

(sizing later tonight)

// indicates commentary

// the following symbols are only used to explain how the notation works and don't actually appear in the notation or its reductions

<> 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)

¡#!,¡%!, ¡@! are separators. What they do is basically separate two numbers.

@ Indicates a cascade. for an example see step 5

// The following are the actual working symbols

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

¡! indicates a separator

{} indicates grouping

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}>...@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:**

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

(sizing later tonight)

- 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!

And now I'll beat Deedlit.

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

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

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

Note that in this:

{x[0]0[0]0}

{x[0]x[0]0[0]0}

{x[0]x[0]x[0]0[0]0}

Never move.

So those xs can follow the known rules.

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

{0,0-0[0]0} = <{0->-0[0]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-x-x[0]0} Can follow the known rules

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

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

And you keep adding higher precedence [0]s and -s in this manner.

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

...

And now, I'll claim that {0[0,0,0,0,0,0,0,0,0,0]0} beats Deedlit.

So first, I have to answer the question "to what does it reduce to?"

And let's look at it upside down. It'll eventually reduce to {0[0,0,0,0,0,0,0,0,0]0}, so we will see to how {0[0,0,0,0,0,0,0,0,0]0} continues.

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},{X}

X follows the known rules (i.e. X can contain any copy of {{0[0,0,0,0,0,0,0,0,0]0},{X}} itself)

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

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

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

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

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

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

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

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

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

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

So:

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

Send:

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

For:

ψ_0(ψ_10(Ω_10))(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}

Note that in this:

{x[0]0[0]0}

{x[0]x[0]0[0]0}

{x[0]x[0]x[0]0[0]0}

Never move.

So those xs can follow the known rules.

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

{0,0-0[0]0} = <{0->-0[0]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-x-x[0]0} Can follow the known rules

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

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

And you keep adding higher precedence [0]s and -s in this manner.

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

...

And now, I'll claim that {0[0,0,0,0,0,0,0,0,0,0]0} beats Deedlit.

So first, I have to answer the question "to what does it reduce to?"

And let's look at it upside down. It'll eventually reduce to {0[0,0,0,0,0,0,0,0,0]0}, so we will see to how {0[0,0,0,0,0,0,0,0,0]0} continues.

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},{X}

X follows the known rules (i.e. X can contain any copy of {{0[0,0,0,0,0,0,0,0,0]0},{X}} itself)

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

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

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

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

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

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

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

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

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

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

So:

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

Send:

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

For:

ψ_0(ψ_10(Ω_10))(3)

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

Vytron wrote: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},<{O1 - a -> 0 <-O2}>

Could this be a typo? I don't know what {O1 - a -> 0 <-O2} means.

case 5: a and b both end in ,0

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

I don't think these are right - b should be the same variable that we used for the other rules, not the last variable of a particular "-" chain.

Send:

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

For:

ψ_0(ψ_10(Ω_10))(3)

Eh? I don't know how you got that.

We have {0,0-0} = LVO = ψ (Ω^Ω^Ω)

X, {0,0-0} = ψ(Ω^(Ω^Ω a))

{0,0-0,0} = ψ(Ω^Ω^(Ω+1))

{0,0,0-0} = ψ(Ω^Ω^(Ω2))

{X-0} = ψ(Ω^Ω^(Ωa))

{0-0-0} = ψ(Ω^Ω^(Ω^2))

{0[0]0} = ψ(Ω^Ω^(Ω^w))

{0[X]0} = ψ(Ω^Ω^(Ω^(wa)))

{0[0,0]0} = ψ(Ω^Ω^(Ω^Ω))

{0[0,0,0]0} = ψ(Ω^Ω^(Ω^Ω^2))

{0[0,0,0,0,0,0,0,0,0,0]0} = ψ(Ω^Ω^(Ω^Ω^9))

So you haven't reached the Bachmann-Howard ordinal yet.

- 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, things are getting fuzzy again. This path may not be worth it exploration anyway if it's this weak, so, huh, let's assume for a moment that I have some working rules to reach this growth (how doesn't matter yet.)

Okay, I had thought that once I managed to reach ψ(Ω^Ω^x) I could just plug any ordinal less than ψ(Ω^Ω^w) in x, and get ψ(Ω^Ω^x) where x is that ordinal.

So, um, I should be able to hold rules strong enough that, say:

{X[0]-0[0]0}

Holds that ordinal, so:

{{{0-0-0-0-0-0-0-0-0}}[0]-0[0]0}

Reaches ψ(Ω^Ω^(Ω^Ω^9))

(In other words, there's an alternative notation that has {0[0,0]0} = {0<-0>[0]0}, and presumably, in that one I can reach ψ(Ω^Ω^(Ω^Ω^9)) at {0[{0-0-0-0-0-0-0-0-0}]0}, can I not, so {{{0-0-0-0-0-0-0-0-0}}[0]-0[0]0} should be able to reach it.)

Is that possible?

And the second question would be, does it matter?

That is, if:

{0[X]0} reaches ψ(Ω^Ω^x) then all I need to do here is having the number eat itself, i.e.

{0[{0,0-0}]0} would reach ψ(Ω^Ω^(Ω^Ω^Ω))

{0{0[{0,0-0}]0}0} would reach ψ(Ω^Ω^(Ω^Ω^(Ω^Ω^Ω)))

So something like:

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

Would reach ψ(ε_{Ω+1})??

Deedlit wrote:We have {0,0-0} = LVO = ψ (Ω^Ω^Ω)

X, {0,0-0} = ψ(Ω^(Ω^Ω a))

{0,0-0,0} = ψ(Ω^Ω^(Ω+1))

{0,0,0-0} = ψ(Ω^Ω^(Ω2))

{X-0} = ψ(Ω^Ω^(Ωa))

{0-0-0} = ψ(Ω^Ω^(Ω^2))

{0[0]0} = ψ(Ω^Ω^(Ω^w))

{0[X]0} = ψ(Ω^Ω^(Ω^(wa)))

{0[0,0]0} = ψ(Ω^Ω^(Ω^Ω))

{0[0,0,0]0} = ψ(Ω^Ω^(Ω^Ω^2))

{0[0,0,0,0,0,0,0,0,0,0]0} = ψ(Ω^Ω^(Ω^Ω^9))

So you haven't reached the Bachmann-Howard ordinal yet.

Okay, I had thought that once I managed to reach ψ(Ω^Ω^x) I could just plug any ordinal less than ψ(Ω^Ω^w) in x, and get ψ(Ω^Ω^x) where x is that ordinal.

So, um, I should be able to hold rules strong enough that, say:

{X[0]-0[0]0}

Holds that ordinal, so:

{{{0-0-0-0-0-0-0-0-0}}[0]-0[0]0}

Reaches ψ(Ω^Ω^(Ω^Ω^9))

(In other words, there's an alternative notation that has {0[0,0]0} = {0<-0>[0]0}, and presumably, in that one I can reach ψ(Ω^Ω^(Ω^Ω^9)) at {0[{0-0-0-0-0-0-0-0-0}]0}, can I not, so {{{0-0-0-0-0-0-0-0-0}}[0]-0[0]0} should be able to reach it.)

Is that possible?

And the second question would be, does it matter?

That is, if:

{0[X]0} reaches ψ(Ω^Ω^x) then all I need to do here is having the number eat itself, i.e.

{0[{0,0-0}]0} would reach ψ(Ω^Ω^(Ω^Ω^Ω))

{0{0[{0,0-0}]0}0} would reach ψ(Ω^Ω^(Ω^Ω^(Ω^Ω^Ω)))

So something like:

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

Would reach ψ(ε_{Ω+1})??

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

Vytron wrote:

Okay, I had thought that once I managed to reach ψ(Ω^Ω^x) I could just plug any ordinal less than ψ(Ω^Ω^w) in x, and get ψ(Ω^Ω^x) where x is that ordinal.

So, um, I should be able to hold rules strong enough that, say:

{X[0]-0[0]0}

Holds that ordinal, so:

{{{0-0-0-0-0-0-0-0-0}}[0]-0[0]0}

Reaches ψ(Ω^Ω^(Ω^Ω^9))

It looks like you are forgetting the ψ. If you have a set of notations that reaches ψ(Ω^Ω^x), and you plug in a notation that reaches ψ(Ω^Ω^9), the ordinal that you reach will be ψ(Ω^Ω^ ψ(Ω^Ω^9)). Diagonalizing over x gets you to the larger ordinal ψ(Ω^Ω^Ω). (That's basically the function of the Ω.)

- 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.

Now, I've realized it's really hard to maintain my notation when I have a series of rules that keep changing depending on the place they are (eg I keep reducing 0,{0},{0,0} to <{0}<,{0}>{0,0}> on a {X} at a point much stronger rules are available.)

So I'm wondering if I can reach the SVO by the time I'm reaching e_0 (I know this ends being an offset, but this is in my interest for the progression's maintenance.)

For this I'll try to have my base notation behave like Deedlit's old box notation, that is, n is ever increasing, so while it increases, so do the number of copies generated.

Begin at the base with:

<x> generates xxx for n+1 xs

n, = (n+1)<+(n+1)> (to match my old base)

n,X,0 = <(>n+1<),X>

Growth:

n,{0} = <(>n+1<)<,0>>

Growth:

And once I have that in place, I can implement the known rules, adding <,0> at the end of everything so we start nesting at the very next step (because I have two kinds of 0s, the ,0s that nest the function, and the {0}s that follow the rules):

These rules apply whenever the number doesn't end in ,0

X stands for an arbitrary sequence {x1},{x},...,{xn}

O stands for an arbitrary sequence of zeroes {0},{0},...,{0}

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

then reduce it: n,X,{a},O -- = n+1,X,{a--},O,<,0>

case 2: all the variables are {0}

n,{0},O -- = = <n+1,{><,O}><,0>

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

n,{0},O,{a,0}} = <n+1,{>{{0},O,{a}}><,O}><,0>

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,O1,{a,0},{0},O2 -- = <n+1,{O1,{a},>{0}<,O2}><,0>

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

n,O1,{a,0},O2,{b,0} = <n+1,{O1,{a},>{n+1{O1,{a,0},O2,{b}}}<,O2}><,0>

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

n,O1,{a},O2,{b,0} = n+1,{O1,{a--},{n+1,{O1,{a},O2,{b}}},O2}<,0>

Would now, something like n,0,{0} = n+1<,{0}><,0> reach the SVO?

Now, I've realized it's really hard to maintain my notation when I have a series of rules that keep changing depending on the place they are (eg I keep reducing 0,{0},{0,0} to <{0}<,{0}>{0,0}> on a {X} at a point much stronger rules are available.)

So I'm wondering if I can reach the SVO by the time I'm reaching e_0 (I know this ends being an offset, but this is in my interest for the progression's maintenance.)

For this I'll try to have my base notation behave like Deedlit's old box notation, that is, n is ever increasing, so while it increases, so do the number of copies generated.

Begin at the base with:

<x> generates xxx for n+1 xs

n, = (n+1)<+(n+1)> (to match my old base)

n,X,0 = <(>n+1<),X>

Growth:

**Spoiler:**

n,{0} = <(>n+1<)<,0>>

Growth:

**Spoiler:**

And once I have that in place, I can implement the known rules, adding <,0> at the end of everything so we start nesting at the very next step (because I have two kinds of 0s, the ,0s that nest the function, and the {0}s that follow the rules):

These rules apply whenever the number doesn't end in ,0

X stands for an arbitrary sequence {x1},{x},...,{xn}

O stands for an arbitrary sequence of zeroes {0},{0},...,{0}

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

then reduce it: n,X,{a},O -- = n+1,X,{a--},O,<,0>

case 2: all the variables are {0}

n,{0},O -- = = <n+1,{><,O}><,0>

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

n,{0},O,{a,0}} = <n+1,{>{{0},O,{a}}><,O}><,0>

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,O1,{a,0},{0},O2 -- = <n+1,{O1,{a},>{0}<,O2}><,0>

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

n,O1,{a,0},O2,{b,0} = <n+1,{O1,{a},>{n+1{O1,{a,0},O2,{b}}}<,O2}><,0>

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

n,O1,{a},O2,{b,0} = n+1,{O1,{a--},{n+1,{O1,{a},O2,{b}}},O2}<,0>

Would now, something like n,0,{0} = n+1<,{0}><,0> reach the SVO?

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

Sorry for the late reply.

This seems suboptimal - better would be

n,{0},{0},O -- = n+1, <{>{{0},O}<,O}><,0>

I think what you have written doesn't work - you have n+1 inserted throughout your notation when it should only be at the beginning.

I believe it should be

n,O1,{a,0},{0},O2 -- = n+1,<{O1,{a},>{0}<,O2}><,0>

The same corrections should be made for cases 3, 5 and 6.

With the above changes, yes it would.

Vytron wrote:case 2: all the variables are {0}

n,{0},O -- = = <n+1,{><,O}><,0>

This seems suboptimal - better would be

n,{0},{0},O -- = n+1, <{>{{0},O}<,O}><,0>

case 4: b = {0}

n,O1,{a,0},{0},O2 -- = <n+1,{O1,{a},>{0}<,O2}><,0>

I think what you have written doesn't work - you have n+1 inserted throughout your notation when it should only be at the beginning.

I believe it should be

n,O1,{a,0},{0},O2 -- = n+1,<{O1,{a},>{0}<,O2}><,0>

The same corrections should be made for cases 3, 5 and 6.

Would now, something like n,0,{0} = n+1<,{0}><,0> reach the SVO?

With the above changes, yes it would.

- 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 see, thanks!

Rewritten rules:

<x> generates xxx for n+1 xs

n, = (n+1)<+(n+1)> (to match my old base)

n,X,0 = <(>n+1<),X>

n,{0} = <(>n+1<)<,0>>

These rules apply whenever the number doesn't end in ,0

X stands for an arbitrary sequence {x1},{x},...,{xn}

O stands for an arbitrary sequence of zeroes {0},{0},...,{0}

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

then reduce it: n,X,{a},O -- = n+1,X,{a--},O,<,0>

case 2: all the variables are {0}

n,{0},{0},O -- = n+1,<{>{{0},O}<,O}><,0>

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

n,{0},O,{a,0}} = n+1,<{>{0},O,{a}}<,O}><,0>

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,O1,{a,0},{0},O2 -- = n+1,<{O1,{a},>{0}<,O2}><,0>

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

n,O1,{a,0},O2,{b,0} = n+1<,{O1,{a},>{{O1,{a,0},O2,{b}}}<,O2}><,0>

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

n,O1,{a},O2,{b,0} = n+1,{O1,{a--},<{>O1,{a},O2,{b}<}>,O2}<,0>

n,0,{0} = n+1<,{0}><,0>

n,{X,0},0,{0} = n+1,{X},0,<,{0}><,0> = phi(a@w)

n,0,0,{0} = n+1,<{>0<,{0}}> = phi(a@w+1)

n,{0-0} = n+1<,0>{0} = phi(1@2w)

n,{0-0},{0-0} = n+1,0<,0>{0-0} = phi(1@3w)

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

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

{0-a} = phi(1@a)

{0-{0,{0}}} = phi(1@phi(1@w))

{0-{0-{0,{0}}} = phi(1@phi(1@phi(1@w)))

{0-{0-{0-{0,{0}}}} = phi(1@phi(1@phi(1@phi(1@w))))

n,{0,0-0},{0} = <{0->{0,{0}}<}> = ψ(Ω^Ω^Ω)(n)

n,{0,0-0},X follows the known rules.

n,X,{0,0-0} = ψ(Ω^(Ω^Ω a))

n,{0,0-0,0}=n<{0->{{0,0-0}}<}><,{0->{{0,0-0}}<}> = ψ(Ω^Ω^(Ω+1))

n,{0[0]0} = n,{0<-0>} = ψ(Ω^Ω^(Ω^w))

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

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

n,{0[0,0]0} = n,{0<[0]0>} = ψ(Ω^Ω^(Ω^Ω))

{0[0,0,0]0} = ψ(Ω^Ω^(Ω^Ω^2))

{0[0,0,0,0,0,0,0,0,0,0]0} = ψ(Ω^Ω^(Ω^Ω^9))

{0[X]0} = ψ(Ω^Ω^(Ω^Ω^a))

{0-0[0]0} = <{0[>X<]0}> ψ(Ω^Ω^(Ω^Ω^Ω))

n,{0-0[0,0]0} = ψ(Ω^Ω^(Ω^Ω^(Ω+1)))

n,{0-0[{0-0[0]0}]0} = ψ(Ω^Ω^(Ω^Ω^Ω^2)))

n,{0,0-0[0]0} = n,{0-<0[{0->X<}]>0} = ψ(Ω^Ω^(Ω^Ω^Ω^w)))

n,{X,0-0[0]0} = n,{X-<0[{X->X<}]0} = ψ(Ω^Ω^(Ω^Ω^Ω^Ω)))

n,{0[0]0-0[0]0} = n<,{>X<-0[0]0}> = ψ(ε_{Ω+1})

(Presumably, since ψ(ε_{Ω+1}) goes ψ(Ω), ψ(Ω^Ω), ψ(Ω^Ω^Ω)... and mine goes ψ(Ω^Ω^Ω), ψ(Ω^Ω^(Ω^Ω)), ψ(Ω^Ω^(Ω^Ω^Ω))...

Rewritten rules:

<x> generates xxx for n+1 xs

n, = (n+1)<+(n+1)> (to match my old base)

n,X,0 = <(>n+1<),X>

n,{0} = <(>n+1<)<,0>>

These rules apply whenever the number doesn't end in ,0

X stands for an arbitrary sequence {x1},{x},...,{xn}

O stands for an arbitrary sequence of zeroes {0},{0},...,{0}

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

then reduce it: n,X,{a},O -- = n+1,X,{a--},O,<,0>

case 2: all the variables are {0}

n,{0},{0},O -- = n+1,<{>{{0},O}<,O}><,0>

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

n,{0},O,{a,0}} = n+1,<{>{0},O,{a}}<,O}><,0>

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,O1,{a,0},{0},O2 -- = n+1,<{O1,{a},>{0}<,O2}><,0>

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

n,O1,{a,0},O2,{b,0} = n+1<,{O1,{a},>{{O1,{a,0},O2,{b}}}<,O2}><,0>

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

n,O1,{a},O2,{b,0} = n+1,{O1,{a--},<{>O1,{a},O2,{b}<}>,O2}<,0>

n,0,{0} = n+1<,{0}><,0>

n,{X,0},0,{0} = n+1,{X},0,<,{0}><,0> = phi(a@w)

n,0,0,{0} = n+1,<{>0<,{0}}> = phi(a@w+1)

n,{0-0} = n+1<,0>{0} = phi(1@2w)

n,{0-0},{0-0} = n+1,0<,0>{0-0} = phi(1@3w)

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

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

{0-a} = phi(1@a)

{0-{0,{0}}} = phi(1@phi(1@w))

{0-{0-{0,{0}}} = phi(1@phi(1@phi(1@w)))

{0-{0-{0-{0,{0}}}} = phi(1@phi(1@phi(1@phi(1@w))))

n,{0,0-0},{0} = <{0->{0,{0}}<}> = ψ(Ω^Ω^Ω)(n)

n,{0,0-0},X follows the known rules.

n,X,{0,0-0} = ψ(Ω^(Ω^Ω a))

n,{0,0-0,0}=n<{0->{{0,0-0}}<}><,{0->{{0,0-0}}<}> = ψ(Ω^Ω^(Ω+1))

n,{0[0]0} = n,{0<-0>} = ψ(Ω^Ω^(Ω^w))

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

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

n,{0[0,0]0} = n,{0<[0]0>} = ψ(Ω^Ω^(Ω^Ω))

{0[0,0,0]0} = ψ(Ω^Ω^(Ω^Ω^2))

{0[0,0,0,0,0,0,0,0,0,0]0} = ψ(Ω^Ω^(Ω^Ω^9))

{0[X]0} = ψ(Ω^Ω^(Ω^Ω^a))

{0-0[0]0} = <{0[>X<]0}> ψ(Ω^Ω^(Ω^Ω^Ω))

n,{0-0[0,0]0} = ψ(Ω^Ω^(Ω^Ω^(Ω+1)))

n,{0-0[{0-0[0]0}]0} = ψ(Ω^Ω^(Ω^Ω^Ω^2)))

n,{0,0-0[0]0} = n,{0-<0[{0->X<}]>0} = ψ(Ω^Ω^(Ω^Ω^Ω^w)))

n,{X,0-0[0]0} = n,{X-<0[{X->X<}]0} = ψ(Ω^Ω^(Ω^Ω^Ω^Ω)))

n,{0[0]0-0[0]0} = n<,{>X<-0[0]0}> = ψ(ε_{Ω+1})

(Presumably, since ψ(ε_{Ω+1}) goes ψ(Ω), ψ(Ω^Ω), ψ(Ω^Ω^Ω)... and mine goes ψ(Ω^Ω^Ω), ψ(Ω^Ω^(Ω^Ω)), ψ(Ω^Ω^(Ω^Ω^Ω))...

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

Vytron wrote:

n,{X,0},0,{0} = n+1,{X},0,<,{0}><,0> = phi(a@w)

n,0,0,{0} = n+1,<{>0<,{0}}> = phi(a@w+1)

This doesn't quite work, as n,0,0,{0} is diagonalized over over {X, {0}} which is a lower level operation. It will be silightly bigger than phi(1@w), but not much (something like phi(2,(phi(1@w)+1))

n,{0-0} = n+1<,0>{0} = phi(1@2w)

n,{0-0},{0-0} = n+1,0<,0>{0-0} = phi(1@3w)

These also need to be fixed somehow.

n,{0,0-0},{0} = <{0->{0,{0}}<}> = ψ(Ω^Ω^Ω)(n)

n,{0,0-0},X follows the known rules.

n,X,{0,0-0} = ψ(Ω^(Ω^Ω a))

n,{0,0-0,0}=n<{0->{{0,0-0}}<}><,{0->{{0,0-0}}<}> = ψ(Ω^Ω^(Ω+1))

These seem to be correct.

n,{0[0]0} = n,{0<-0>} = ψ(Ω^Ω^(Ω^w))

That should be n,{0[0]0} = n,{0<-0>} = ψ(Ω^Ω^(Ωw))

n,{0[0,0]0} = n,{0<[0]0>} = ψ(Ω^Ω^(Ωw^2))

{0[0,0,0]0} = ψ(Ω^Ω^(Ωw^3))

{0[0,0,0,0,0,0,0,0,0,0]0} = ψ(Ω^Ω^(Ωw^9))

Beyond this point I need more clarity on how the rules work, but I'm pretty sure this is what we get:

{0[X]0} = ψ(Ω^Ω^(Ω^a))

{0-0[0]0} = <{0[>X<]0}> ψ(Ω^Ω^(Ω^Ω))

n,{0-0[0,0]0} = ψ(Ω^Ω^((Ω^Ω)+1))

n,{0-0[{0-0[0]0}]0} = ψ(Ω^Ω^((Ω^Ω)+ψ(Ω^Ω^(Ω^Ω)))))

n,{0,0-0[0]0} = n,{0-<0[{0->X<}]>0} = ψ(Ω^Ω^((Ω^Ω)2))

n,{X,0-0[0]0} = n,{X-<0[{X->X<}]0} = ψ(Ω^Ω^((Ω^Ω)a))

n,{0[0]0-0[0]0} = n<,{>X<-0[0]0}> =ψ(Ω^Ω^(Ω^(Ω2)))

- 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...

Now, I'll admit defeat. Looks like creating a bottom-top notation only can get one so far, because by the next hierarchy it didn't matter how one jumped at the base (everything before then becomes an offset) and it's all about the rules that you use by then, and all amount of weaker rules and extensions to them are going to be below some limit ordinal, which is using stronger rules.

I spent a lot of time developing rules that got me to e_0, by the time I was at Gamma_0 they were useless (rules that only went to w^w^w would have worked there), and by the time I reached the SVO, the rules of getting to Gamma_0 were useless too (the set of rules that got me to the SVO could have been used from the start), and any previous set of rules seem to keep me under ψ(Ω^Ω^Ω^Ω^Ω).

Which made my fears here valid:

I see no other option, so:

Let:

1 = 0,0

2 = 0,0,0

3 = 0,0,0,0

4 = 0,0,0,0,0

...

I.e. no integers ever appear, n+1 adds a 0 and m-1 removes a 0.

-- = "Reduce this"

<> = "Copy this"

n,{0-} = n

n,{0-X} = n+1,{0- X--}

Rules for Y in n,{0-Y} (i.e. these rules only apply when Y is preceded by n,{0- at the very beginning)

X-{0} -- = X

A-{B-{0}- m}-D -- = A<-{B- m}>-D, where D does not contain any }'s

A-{B-{m}-C- n}-D -- = A-{<B{><m-1}-C> n}-D, where n < m, C and D do not contain any }'s and C does not contain any x} with x < m.

n,{0,0-} = n+1,{0-<{><n}->0<}>

Send:

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

Which expands:

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

Now, I'll admit defeat. Looks like creating a bottom-top notation only can get one so far, because by the next hierarchy it didn't matter how one jumped at the base (everything before then becomes an offset) and it's all about the rules that you use by then, and all amount of weaker rules and extensions to them are going to be below some limit ordinal, which is using stronger rules.

I spent a lot of time developing rules that got me to e_0, by the time I was at Gamma_0 they were useless (rules that only went to w^w^w would have worked there), and by the time I reached the SVO, the rules of getting to Gamma_0 were useless too (the set of rules that got me to the SVO could have been used from the start), and any previous set of rules seem to keep me under ψ(Ω^Ω^Ω^Ω^Ω).

Which made my fears here valid:

Vytron wrote: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...

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.

I see no other option, so:

Let:

1 = 0,0

2 = 0,0,0

3 = 0,0,0,0

4 = 0,0,0,0,0

...

I.e. no integers ever appear, n+1 adds a 0 and m-1 removes a 0.

-- = "Reduce this"

<> = "Copy this"

n,{0-} = n

n,{0-X} = n+1,{0- X--}

Rules for Y in n,{0-Y} (i.e. these rules only apply when Y is preceded by n,{0- at the very beginning)

X-{0} -- = X

A-{B-{0}- m}-D -- = A<-{B- m}>-D, where D does not contain any }'s

A-{B-{m}-C- n}-D -- = A-{<B{><m-1}-C> n}-D, where n < m, C and D do not contain any }'s and C does not contain any x} with x < m.

n,{0,0-} = n+1,{0-<{><n}->0<}>

Send:

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

Which expands:

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

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

Vytron wrote:

Now, I'll admit defeat. Looks like creating a bottom-top notation only can get one so far, because by the next hierarchy it didn't matter how one jumped at the base (everything before then becomes an offset) and it's all about the rules that you use by then, and all amount of weaker rules and extensions to them are going to be below some limit ordinal, which is using stronger rules.

I spent a lot of time developing rules that got me to e_0, by the time I was at Gamma_0 they were useless (rules that only went to w^w^w would have worked there), and by the time I reached the SVO, the rules of getting to Gamma_0 were useless too (the set of rules that got me to the SVO could have been used from the start), and any previous set of rules seem to keep me under ψ(Ω^Ω^Ω^Ω^Ω).

Well, it does seem like you could keep on going up to the Bachmann-Howard ordinal, you just don't reach it as soon as you were expecting. But each nesting of brackets could get you another exponent or two of Ω, so the limit of nesting should be the BHO.

1 = 0,0

2 = 0,0,0

3 = 0,0,0,0

4 = 0,0,0,0,0

...

I.e. no integers ever appear, n+1 adds a 0 and m-1 removes a 0.

-- = "Reduce this"

<> = "Copy this"

n,{0-} = n

n,{0-X} = n+1,{0- X--}

Rules for Y in n,{0-Y} (i.e. these rules only apply when Y is preceded by n,{0- at the very beginning)

X-{0} -- = X

A-{B-{0}- m}-D -- = A<-{B- m}>-D, where D does not contain any }'s

A-{B-{m}-C- n}-D -- = A-{<B{><m-1}-C> n}-D, where n < m, C and D do not contain any }'s and C does not contain any x} with x < m.

Well, that looks familiar!

n,{0,0-} = n+1,{0-<{><n}->0<}>

Send:

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

Which expands:

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

[/quote]

That's smaller than my number. Of course, we have the same notation now, so we could easily play an infinite game of one-upsmanship here.

- 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!

That's smaller than my number.

Interesting, I've gotten so used to nesting being stronger than bigger input, that I expected nesting a smaller number more times would bring a bigger number than yours. But now I see your number would eventually reduce to n | (((3)3)3)0) for some huge n...

What bothers me, is that up to introducing []s on my old notation, it and the one I just defined use the same symbols (you have some n followed by ,s, and stuff nested in {}s, separated by -s), but the latter is mindbogglingly much stronger than the former, despite the latter not having many definitions for symbols (I don't think the latter has some reduction rules for something like 0,0,{0-{0,{0-{0,{0,0},0}-0,{0},0}-0,0,{0-0}}-0}}}} so some extension could include stronger definitions to collapse it into something even stronger), so the latter is superior to the former in all ways, and it's more extensible (while the former reached its limits of extensibility and that's why I needed to introduce the []s)

Working on extending the old notation feels like getting a faster horse or better tires for my carriage when trying to beat a rocket.

But, if the alternative is a boring one-upsmanship game, then I guess...

Deedlit wrote:Vytron wrote:

n,{X,0},0,{0} = n+1,{X},0,<,{0}><,0> = phi(a@w)

n,0,0,{0} = n+1,<{>0<,{0}}> = phi(a@w+1)

This doesn't quite work, as n,0,0,{0} is diagonalized over over {X, {0}} which is a lower level operation. It will be silightly bigger than phi(1@w), but not much (something like phi(2,(phi(1@w)+1))n,{0-0} = n+1<,0>{0} = phi(1@2w)

n,{0-0},{0-0} = n+1,0<,0>{0-0} = phi(1@3w)

These also need to be fixed somehow.n,{0,0-0},{0} = <{0->{0,{0}}<}> = ψ(Ω^Ω^Ω)(n)

n,{0,0-0},X follows the known rules.

n,X,{0,0-0} = ψ(Ω^(Ω^Ω a))

n,{0,0-0,0}=n<{0->{{0,0-0}}<}><,{0->{{0,0-0}}<}> = ψ(Ω^Ω^(Ω+1))

These seem to be correct.

Correct despite weaknesses on early recursions? So they don't seem to matter?

It's curious, I switched to a much stronger notation that allowed me to reach the SVO much earlier, but the reduction rules used right after that, turn out to be weaker and I actually have to catch up with my old notation and it's possible n,{0,0-0,0} of my old notation produced a slightly bigger number than n,{0,0-0,0} on the new one.

Perhaps it wasn't a good idea to drop mike's notation after all...

Let's go hybrid:

<x> generates xxx for n+1 xs

n, = (n+1)<+(n+1)>

n,X,0 = <(>n+1<),X>

n,{0} = <(>n+1<)<,0>>

When X does not contain any -:

**Spoiler:**

n,{0-0} = <(>n+1<)<{>0<}>>

When X does contain any -:

**Spoiler:**

n,0,{0-0} = n+1,{0-0}, {{0}<,{0}>}<,0>

n,{X,0},{0-0} = n+1,{X},{0-0},{{0}<,{0}>}<,0>

n,0,0,{0-0} = n+1,<{>0<,{0-0}}><,0>

n,{0-0},{0-0} = n+1<,0>{0-0}<,0>

n,{0-0},{0-0},{0-0} = n+1<,0>{0-0},{0-0}<,0>

n,{0-0,0} = n+1,{0-0}<,{0-0}><,0>

n,{0-0,0,0} = n+1,{0-0,0}<,{0-0,0}><,0>

n,{0,0-0} = n+1,<{0->{{0-0},{0-0}}<}><,0>

n,{0[0]0} = n+1,{0<-0>}<,0>

n,0,{0[0]0} = n+1,{0[0]0}, {{0}<,{0}>}<,0>

n,{X,0},{0[0]0} = n+1,{X},{0[0]0},{{0}<,{0}>}<,0>

n,0,0,{0[0]0} = n+1,<{>0<,{0[0]0}}><,0>

n,{0[0]0},{0[0]0} = n+1<,0>{0[0]0}<,0>

n,{0[0]0},{0[0]0},{0[0]0} = n+1<,0>{0[0]0},{0[0]0}<,0>

n,{0[0]0,0} = n+1,{0[0]0}<,{0[0]0}><,0>

n,{0,0[0]0} = n+1,<{0[0]>{{0[0]0},{0[0]0}}<}><,0>

n,{0[0]0-0} = n+1,{0<[0]0>}<,0>

n,{0[0]0-0-0} = n+1,{0<[0]0-0>}<,0>

n,{0[0]0-0-0-0} = n+1,{0<[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} = n+1,{0<[0,0]0-0>}<,0>

n,{0[0,0]0-0-0-0} = n+1,{0<[0,0]0-0-0>}<,0>

n,{0[0,0]0[0]0} = n+1,{0[0,0]0<-0>}<,0>

n,{0[0,0]0[0]0-0} = n+1,{0[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>

n,{0[0,0,0]0} = n+1,{0[0,0]0[0]0<-0>}<,0>

Now, the idea here is to add prefixes.

So we'd have:

n,{0-0[0]0} = n+1,<{0[>{0[0]0},{0[0]0}<]0}><,0>

Because at no point a - was used as a separator before a [0] separator.

And now each prefix counts how many times the previous prefixes counted up:

n,{0,0-0[0]0} = n+1,<{0-0[>{0-0[0]0},{0-0[0]0}<]0}><,0>

And those prefixes follow the known rules up until:

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

Is this still at ψ(Ω^Ω^(Ω^(Ω2)))?

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

I challenge all your notations on the grounds that you must ask for sizings and corrections

according to rule 4 your number should be easy to understand for the layman.

a layman will surely understand the bits and details of a notation to a degree at least below the creator

so if you cant proper understand all the implications of your work, it follows that a layman will also be unable to do so.

according to rule 4 your number should be easy to understand for the layman.

a layman will surely understand the bits and details of a notation to a degree at least below the creator

so if you cant proper understand all the implications of your work, it follows that a layman will also be unable to do so.

- 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 challenge your challenge on the grounds that it doesn't specify against what number your challenge applies, because I haven't sent a number yet with this.

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

Challenge accepted: You failed to properly read. I said: "I challenge ALL YOUR NOTATIONS" so any number that uses any notation sent by you previous to my challenge is held challenged. (This includes several numbers you sent with notations that have been already dropped).

Since your challenge to my challenge has been satisfied, my challenge stands, removing all your notations entirely from the competition

Since your challenge to my challenge has been satisfied, my challenge stands, removing all your notations entirely from the competition

- 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 claim that I understand the bits and details of at least one of my notations, therefore making your claim invalid (you'd need to show that I don't properly understand all the implications of ALL my notations to take them all down.)

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

"I claim that I understand the bits and details of at least one of my notations"

The burden of proof is on you for this, i don't need to show that you don't, you need to show me that you do.

The burden of proof is on you for this, i don't need to show that you don't, you need to show me that you do.

- 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, let's take a look at a notation I sent in this post:

In this notation, you have a series of 1, such that:

0 = 0

1 = 1

11 = 2

111 = 3

1111 = 4

I understand the bits and details of it since simply, to know the value of a random string such as:

1111111111111111111

All you have to do is add a + between each 1, and calculate a result, i.e.:

1111111111111111111 = 1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1 = 19

And so on for any number of 1s.

QED

Since your challenge applies to ALL MY NOTATIONS it includes this one. Since I understand it, the layman understands it, your premise is false and your challenge falls down.

Vytron wrote:the rules allow for someone to claim that they don't understand how 11 is bigger then 10, presumably by then you'd explain how numbers work and that in unary:

11111111111 > 1111111111

Because it has one term more.

And until you do that, your 11 would be invalid..

In this notation, you have a series of 1, such that:

0 = 0

1 = 1

11 = 2

111 = 3

1111 = 4

I understand the bits and details of it since simply, to know the value of a random string such as:

1111111111111111111

All you have to do is add a + between each 1, and calculate a result, i.e.:

1111111111111111111 = 1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1 = 19

And so on for any number of 1s.

QED

Since your challenge applies to ALL MY NOTATIONS it includes this one. Since I understand it, the layman understands it, your premise is false and your challenge falls down.

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

Since you did not invent unary, my challenge stands.

But i'll grace you and Challenge again, All your notations except your clone of unary, same grounds as my previous challenge

But i'll grace you and Challenge again, All your notations except your clone of unary, same grounds as my previous challenge

- 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!

Nope, unary doesn't have a 0, my notation does have a 0.

"Except your unary clone notation" doesn't exclude that notation that I understand, because for that you'd need to prove it's a clone of unary (which is not, because unary doesn't have a symbol for 0, because it would mean it has at least 2 symbols, and then it'd not be unary anymore.)

"Except your unary clone notation" doesn't exclude that notation that I understand, because for that you'd need to prove it's a clone of unary (which is not, because unary doesn't have a symbol for 0, because it would mean it has at least 2 symbols, and then it'd not be unary anymore.)

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

I Challenge again, All your notations except this one:

same grounds as my previous challenge

In this notation, you have a series of 1, such that:

0 = 0

1 = 1

11 = 2

111 = 3

1111 = 4

same grounds as my previous challenge

- 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'll now claim that I have invented an infinite number of notations, and so that even if you challenged them one by one you'll never be able to challenge them all.

I suggest that instead of trying to create a wide net that covers for everything, you actually list the notations that you claim "I can't properly understand all the implications of" so that you only take down the relevant ones (for instance, I doubt you were interested in taking down the pseudobinary one).

Lest we'll end stuck in a loop where I list one by one all my notations that I understand and you'll have to exclude them one by one, which will take a long while.

I suggest that instead of trying to create a wide net that covers for everything, you actually list the notations that you claim "I can't properly understand all the implications of" so that you only take down the relevant ones (for instance, I doubt you were interested in taking down the pseudobinary one).

Lest we'll end stuck in a loop where I list one by one all my notations that I understand and you'll have to exclude them one by one, which will take a long while.

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

I got time.

I Challenge that on the grounds that it is false. (Provide proof)

"I'll now claim that I have invented an infinite number of notations, and so that even if you challenged them one by one you'll never be able to challenge them all."

I Challenge that on the grounds that it is false. (Provide proof)

- 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!

You didn't provide proof that I don't understand ALL my notations, so why would I need proof to state I have built an infinite number of notations?

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

why would I need proof to state I have built an infinite number of notations?

The burden of proof is on the party who asserts the claim.

I can tell you, i have 5 oranges, prove me wrong. If i show you the oranges as proof, that is my proof and my obligation as the claim-making party is fullfilled.

If i tell you, i have infinity oranges, prove me wrong. There is still an obligation on my part to prove it before i can say "prove me wrong"

- 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!

Daggoth wrote:The burden of proof is on the party who asserts the claim.

Then the burden of proof that I don't understand ALL my notations is on you.

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

Here is my proof for that claim:

you must ask for sizings and corrections

according to rule 4 your number should be easy to understand for the layman.

a layman will surely understand the bits and details of a notation to a degree at least below the creator

so if you cant proper understand all the implications of your work, it follows that a layman will also be unable to do so.

- 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!

You say "if you cant proper understand all the implications of your work, it follows that a layman will also be unable to do so", but you don't show THAT I cant proper understand all the implications of my work.

### Who is online

Users browsing this forum: No registered users and 25 guests