## Largest total computable function competition

### Re: Largest total computable function competition

Oh, so if f(x) = 3x+2 then f^3(5) = 161?

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- Moosey
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### Re: Largest total computable function competition

Wikipedia has an article explaining that it is not to be used as the only source for research.testitemqlstudop wrote:The biggest lie perpetrated by humanity (cough. cough. TEACHERS) is that Wikipedia is unreliable. Studies show [citation needed] that for every 1 edit to a page, ten read the page.Moosey wrote: according to Wikipedia, the incredibly unreliable resource™.

Yes.PkmnQ wrote:Oh, so if f(x) = 3x+2 then f^3(5) = 161?

I am a prolific creator of many rather pathetic googological functions

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

- testitemqlstudop
**Posts:**1323**Joined:**July 21st, 2016, 11:45 am**Location:**in catagolue-
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### Re: Largest total computable function competition

Where on the fgh is my function for, say, a function f with growth rate w^2

- Moosey
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### Re: Largest total computable function competition

Are you asking for œ(f_w^2,n,n)?testitemqlstudop wrote:Where on the fgh is my function for, say, a function f with growth rate w^2

I'm not sure why I ask since I can't calculate it, but I doubt it's too much higher.

Variant of your function:

Code: Select all

```
æ(F, x, y, z) = {
F(y+z) when x = 0
F(æ(F^x, x-1, æ(F^x, x-1, F(x), z+1))+z, z+1) when y = 0
æ(F^y, x, y-1, z+1) otherwise
}
```

Almost at this point, unfortunately:

The googology wiki wrote:[salad numbers] are often coined by inexperienced googologists holding the false notions that a more complex definition is a more powerful one.

I am a prolific creator of many rather pathetic googological functions

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

### Re: Largest total computable function competition

I wonder what æ(S,n,n,n) would be. After all, I did the same with œ.Moosey wrote:Are you asking for œ(f_w^2,n,n)?testitemqlstudop wrote:Where on the fgh is my function for, say, a function f with growth rate w^2

I'm not sure why I ask since I can't calculate it, but I doubt it's too much higher.

Variant of your function:Code: Select all

`æ(F, x, y, z) = { F(y+z) when x = 0 F(æ(F^x, x-1, æ(F^x, x-1, F(x), z+1))+z, z+1) when y = 0 æ(F^y, x, y-1, z+1) otherwise }`

Almost at this point, unfortunately:The googology wiki wrote:[salad numbers] are often coined by inexperienced googologists holding the false notions that a more complex definition is a more powerful one.

æ(S,1,1,1) = æ(S,1,0,2) = æ(S,0,æ(S,0,2,3)+2,3) + 1 = æ(S,0,7,3) + 1 = 10 + 1 = 11

æ(S,2,2,2) = æ(S^2,2,0,4) = æ(S^4,1,æ(S^4,1,4,5)+4,5) + 2 = æ(S^4,1,æ(S^96,1,0,9)+4,5) + 2 = æ(S^4,1,æ(S^96,0,97,10)+100,5) + 2 = æ(S^4,1,207,5) + 2 = æ(S^4*(207!),1,0,212) + 2 = æ(S^4*(207!),0,(4*(207!))+1,213) + 4*(207!)+2 = (8*(207!)) + 216

If you recall, œ(S,2,2) was as low as 195.

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- testitemqlstudop
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### Re: Largest total computable function competition

S is f_w, right?

Manually trying to bash out ae(S,4,4):

ae(S,4,4) =

ae(S^4,4,3) =

ae(S^12,4,2) =

ae(S^24,4,1) =

ae(S^24,4,0)=

ae(S^96,3,28)+24=... oh strewth this is not a good idea

Anyways I think this reduction is obvious:

Hence, applying the Y=0, rule, we get

because of applying the rule œ(F^(y!), x, 0)=F^(y!)(œ(F^(y!*x), x-1, F(x))).

Can someone bash further?

I think it is obvious that for f=S then œ is at least as powerful as iterated factorial.

Code: Select all

```
œ(F, x, y) = {
F(y) when x = 0
F(œ(F^x, x-1, F(x))) when y = 0
œ(F^y, x, y-1) otherwise
}
```

ae(S,4,4) =

ae(S^4,4,3) =

ae(S^12,4,2) =

ae(S^24,4,1) =

ae(S^24,4,0)=

ae(S^96,3,28)+24=... oh strewth this is not a good idea

Anyways I think this reduction is obvious:

Code: Select all

```
œ(F, x, y) = {
F(y) when x = 0
F(œ(F^x, x-1, F(x))) when y = 0
œ(F^(y!), x, 0) otherwise
}
```

Code: Select all

```
œ(F, x, y) = {
F(y) when x = 0
F^(y!)(œ(F^(y!*x), x-1, F(x))) otherwise
}
```

Can someone bash further?

I think it is obvious that for f=S then œ is at least as powerful as iterated factorial.

- Moosey
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### Re: Largest total computable function competition

No, S(n) = n+1testitemqlstudop wrote:S is f_w, right?

If you're asking about whether œ(S,n,n) or æ(S,n,n,n) are f_w, I don't know. Probably not.

Uh, æ has four argumentstestitemqlstudop wrote:

Manually trying to bash out ae(S,4,4):

ae(S,4,4) =

ae(S^4,4,3) =

ae(S^12,4,2) =

ae(S^24,4,1) =

ae(S^24,4,0)=

ae(S^96,3,28)+24=...

æ(S, 4, 4, 4)

= æ(S^4, 4, 3, 5)

= æ(S^12, 4, 2, 6)

= æ(S^24, 4, 0, 8)

= 24 + æ(S^96, 3, æ(S^96, 3, 28, 9))+8, 9)

= 24 + æ(S^96, 3, æ(S^(96*(28!)), 3, 0, 37))+8, 9)

...

I am a prolific creator of many rather pathetic googological functions

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

### Re: Largest total computable function competition

Hmm, I wonder if I can make my own function based off of œ. How about for 2-argument functions?

Code: Select all

```
ø(F, n, x, y) =
F(x,y) [n=0]
F(ø(F^n,n-1,F(n,n),F(n,n)),ø(F^n,n-1,F(n,n),F(n,n))) [x=0, y=0]
F(ø(F^x,n,x-1,F(x,n)),F(ø(F^x,n,x-1,F(x,n)) [y=0]
ø(F^y,n,x,y-1)
```

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- testitemqlstudop
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### Re: Largest total computable function competition

How do you define F^x for multi-variable functions tho

EDIT: Whoops i meant oe

EDIT: Whoops i meant oe

### Re: Largest total computable function competition

Oh, theres no actual description yet?testitemqlstudop wrote:How do you define F^x for multi-variable functions tho

EDIT: Whoops i meant oe

It might be f(f(x, x), f(y, y)) maybe?

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- Moosey
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### Re: Largest total computable function competition

Here's a compacted definition of æ:

For f(x) = 2^x:

æ(f,2,2,2) =

æ(f^2,2,0,4) =

2^(æ(f^4, 1, æ(f^4, 1, 16, 5))+4, 5)

æ(f^4, 1, 16, 5) =

æ(f^83691159552000,1,0,21)= (you're probably thinking, "holy mustard, where did that come from?")

f^83691159552000(æ(f^83691159552000, 0, æ(f^83691159552000, 0, 2^^83691159552000, 22))+21, 22)

æ(f^83691159552000, 0, 2^^83691159552000, 22)= f^83691159552000(2^^83691159552000+22)

f^83691159552000(æ(f^83691159552000, 0, æ(f^83691159552000, 0, 2^^83691159552000, 22))+21, 22) =

f^83691159552000(æ(f^83691159552000, 0, f^83691159552000(2^^83691159552000+22)+21, 22)=

f^167382319104000(f^83691159552000(2^^83691159552000+22)+43))

2^(æ(f^4, 1, æ(f^4, 1, 16, 5))+4, 5) =
etc, etc

EDIT:

Further compacted æ:

Code: Select all

```
æ(F, x, y, z) = {
F(y+z) when x = 0
F(æ(F^x, x-1, æ(F^x, x-1, F(x), z+1))+z, z+1) when y = 0
æ(F^(y!), x, 0, z+y) otherwise
}
```

æ(f,2,2,2) =

æ(f^2,2,0,4) =

2^(æ(f^4, 1, æ(f^4, 1, 16, 5))+4, 5)

æ(f^4, 1, 16, 5) =

æ(f^83691159552000,1,0,21)= (you're probably thinking, "holy mustard, where did that come from?")

f^83691159552000(æ(f^83691159552000, 0, æ(f^83691159552000, 0, 2^^83691159552000, 22))+21, 22)

æ(f^83691159552000, 0, 2^^83691159552000, 22)= f^83691159552000(2^^83691159552000+22)

f^83691159552000(æ(f^83691159552000, 0, æ(f^83691159552000, 0, 2^^83691159552000, 22))+21, 22) =

f^83691159552000(æ(f^83691159552000, 0, f^83691159552000(2^^83691159552000+22)+21, 22)=

f^167382319104000(f^83691159552000(2^^83691159552000+22)+43))

2^(æ(f^4, 1, æ(f^4, 1, 16, 5))+4, 5) =

Code: Select all

`2^(æ(f^4, 1, f^167382319104000(f^83691159552000(2^^83691159552000+22)+43))+4, 5)`

EDIT:

Further compacted æ:

Code: Select all

```
æ(F, x, y, z) =
F(y+z), x = 0
F^(y!)(æ(F^(*y!x), x-1, æ(F^(y!*x), x-1, F^(y!)(x), z+y+1))+z+y, z+y+1), x not = 0
```

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

- testitemqlstudop
**Posts:**1323**Joined:**July 21st, 2016, 11:45 am**Location:**in catagolue-
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### Re: Largest total computable function competition

Here I have specialized oe for S:

... yeah it's pitifully small.

Code: Select all

```
oes(n, x, y) = {
y + n when x = 0
oes(n*y!*x, x-1, x+n) + y! otherwise
}
oes(x) = oes(x, x)
oes(x, y) = oes(1, x, y)
oes(4) = oes(1, 4, 4) =
oes(96, 3, 5) + 24 =
oes(34650, 2, 99) + 144 =
oes(69300*99!, 1, 34652) + 99! + 144 =
oes(69300*99!*34652!, 0, 1+69300*99!) + 34652! + 99! + 144 =
1 + 69300*99! + 69300*99!*34652! + 34652! + 99! + 144 =
69300*(99!*34652! + 1) + 34652! + 99! + 145
```

- Moosey
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### Re: Largest total computable function competition

define tvi(f,n,x,y) (two-var-iteration function) as follows:

tvi(f,n,x,y)=
tvi is essentially this:

S^3^.3*S^3^.3*S^3^.3(1)(1)(1)=
...

Can anyone evaluate it?

tvi(f,n,x,y)=

Code: Select all

```
f(tvi(f,n-1,x,x),tvi(f,n-1,y,y)), n>1
f(x,y), n=1
```

EDIT:PkmnQ wrote:Oh, theres no actual description yet?testitemqlstudop wrote:How do you define F^x for multi-variable functions tho

EDIT: Whoops i meant oe

It might be f(f(x, x), f(y, y)) maybe?

S^3(:1:)3(1) = S^3|3(1) = S^3^.3^.3(1) =A while ago, I wrote:where is S^2||x(1) in the grass-mowing hierarchy? (Probably very low--I mean the fgh of course)A while before that, I wrote:This morning I created a notation that extends f^a(x). I will use ^. to notate an important part of it so it’s not confused with exponentiation, but think of it is superscript right above something.

It starts like this:

f^1^.a(x) = f^f^f^f^f^f^f^f^f...(x)(x)(x)(x)... with a f’s, all of x

so, where f = +1,

f^1^.5(1)=then: f^b^.a(x) = f^b*f^b*f...(x)(x)(x) with a b*f’s.Code: Select all

`f^f^f^f^f(1)(1)(1)(1)(1)= f^f^f^f^2(1)(1)(1)(1) f^f^f^3(1)(1)(1)... =6`

then, where # represents anything, including a stack: f^#^.b^.a(x)=f^#^.b*f^#^.b*f...(x)(x)(x) with a f^#^.b*’s.

then we have this leap:

f^a|n(x)= f^a^.a^.a^.a^.a ... with n a's.

f^a|n^.#(x)=f^(f^#(a))^.(f^#(a))^.(f^#(a)) ... with n (f^#(a))^. ’s. The initial f, is, of course, still of x.

Then, the next logical extension:

f^a|n|m(x) = f^a|n^.a|n^.a|n... with m a|n ’s.

The logical continuation:

f^a|n|m^.#(x)= f^f^a|n^.#(a)^f^a|n^.#(a)^f^a|n^.#(a)... with f^n|m(a) f’s. The initial f will always be of x of course

Can somebody help me define f^a|b|c|d... with arbitrarily many arguments?

EDIT: nvm, f^a|n|m|#(x)= f^a|n|f^#(m)(x)

Then I can, a la bowers, define f^a||b(x) = f^a|a|a|a|a...(x) with b a’s.

Anyways, now:

f^#||b||c(x) = f^#||(f^b||c(b))(x)

GENERALIZATION TIME YAY

f^a(:b:)c(x) =f^#a(:b:)c(x) =Code: Select all

`f^a(:b-1:)a(:b-1:)a(x) with c a's, b > 1 f^a|c, b = 1`

I'm sure that S^3(:x:)3(1) is much higher in the fgh. At least, relatively speaking considering S = the successor function.Code: Select all

`f^#a|c(x), b = 1 f^#(:b:)(f^a(:b:)c(a))(x), b > 1`

S^3^.3*S^3^.3*S^3^.3(1)(1)(1)=

Code: Select all

`S^3^.3*S^3^.66(1)(1)`

Can anyone evaluate it?

Last edited by Moosey on October 3rd, 2019, 4:44 pm, edited 3 times in total.

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

- Moosey
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### Re: Largest total computable function competition

I finally understand the fgh now, and I'm not afraid to use it. First off, a function which eventually dominates f_a(x) for all a < e_0, using the fairly obvious method of... using the fgh

It uses ord, too.

Here's how it works:

ord-fgh(n) =

max{f_ord(1)(n), f_ord(2)(n), f_ord(3)(n)... f_ord(n)(n)}

Provably, at n= 2^^a it suddenly becomes f_w^^a(n), so as long as you have a finite amount of w's in the bottom you can always find an n such that it exceeds f_(the ordinal you're talking about)(n). Therefore it is on par with f_e_0(n), save that it goes much much more slowly than that. But it beats everything below it eventually, so...

The second function (which we'll get to in quite a while) is based on a naïvely created relative of the fgh:

j_n(x) =
Where fin(n,x) is defined thus:

fin(n,x) =
Yes, fin(n,x) returns a finite value as long as x is finite

Finally, we can get to my second function on par with f_e_0

Then I defined ordj (and yes, this was originally ordi):

ordj(n) = max{j_ord(1)(n), j_ord(2)(n), ...j_ord(n)(n)}

and of course eventually dominates everything < j_e_0(n) and still is probably on par with f_e_0

It uses ord, too.

Here's how it works:

ord-fgh(n) =

max{f_ord(1)(n), f_ord(2)(n), f_ord(3)(n)... f_ord(n)(n)}

Provably, at n= 2^^a it suddenly becomes f_w^^a(n), so as long as you have a finite amount of w's in the bottom you can always find an n such that it exceeds f_(the ordinal you're talking about)(n). Therefore it is on par with f_e_0(n), save that it goes much much more slowly than that. But it beats everything below it eventually, so...

The second function (which we'll get to in quite a while) is based on a naïvely created relative of the fgh:

j_n(x) =

Code: Select all

```
(Technically it was i_n(x) but autocorrect is a REAL PAIN
x+1, n=0
j^(x*(fin(n,x)+1))_n-1(x), n > 0 & not a limit ordinal
j^(fin(n,x)+1)_n[x](x), n a limit ord
```

fin(n,x) =

Code: Select all

```
fin(n-1, x)+1, n not a limit ordinal, n>w
n, n < w
fin(n[x], x)+1, n a limit ordinal, n>w
```

Finally, we can get to my second function on par with f_e_0

Then I defined ordj (and yes, this was originally ordi):

ordj(n) = max{j_ord(1)(n), j_ord(2)(n), ...j_ord(n)(n)}

and of course eventually dominates everything < j_e_0(n) and still is probably on par with f_e_0

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

### Re: Largest total computable function competition

I haven’t seen n[x] before.

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- Moosey
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### Re: Largest total computable function competition

n[x]PkmnQ wrote:I haven’t seen n[x] before.

= the xth term of the fundamental sequence of n

so, for instance, w[400] = the 400th term of {0,1,2,3,4,5,6,7,8...} = 400 (at least, this is probably the most formal way to say it-- you could also call it 399 or 401 depending on whether you start your series with entry 0 or 1 or whether omega's sequence includes 0)

and e_0[3] = the 3rd term if {w, w^w, w^w^w, w^w^w^w...} = w^^4 (again, if you start a series with 0 or 1)

So the fgh is actually defined like this:

f_n(x)=

Code: Select all

```
x+1, n = 0
f^x_(n-1)(x), n > 0, n not a limit ordinal
f_n[x](x), n a limit ordinal
```

If you don't know what limit ordinals are, they're distressingly simple. They're just any ordinal you cannot subtract one from. In other words, you aren't adding any finite constants. w, w2, and w^w^w^w +w^3 + w19 are all limit ordinals

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

- testitemqlstudop
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### Re: Largest total computable function competition

So the sequence for w^w is [w, w*w, w*w*w, w*w*w*w, ...] and the sequence for w*w is [w, w*2, w*3, w*4, ...]?

Your explanation is better than the one on goowiki

Your explanation is better than the one on goowiki

- Moosey
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### Re: Largest total computable function competition

Exactlytestitemqlstudop wrote:So the sequence for w^w is [w, w*w, w*w*w, w*w*w*w, ...] and the sequence for w*w is [w, w*2, w*3, w*4, ...]?

Thankstestitemqlstudop wrote:Your explanation is better than the one on goowiki

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

- Moosey
**Posts:**3041**Joined:**January 27th, 2019, 5:54 pm**Location:**A house, or perhaps the OCA board. Or [click to not expand]-
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### Re: Largest total computable function competition

(See here)
Does this do what I want?

Also they added new BBcode smilies which totally screwed some of my notation up ( () ())

Code: Select all

```
ssm_n(x,y)=
x+y, n = 0
ssm_sg(x,n,x,y)(x,y), n a lim ord
ssm_n-1(x,xy) otherwise
define sg(m,n,x,y) =
n[ssm_sg(m-1,n,x)(x,y)], m>0,
x, m less than or = to 0
```

Also they added new BBcode smilies which totally screwed some of my notation up ( () ())

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

- Moosey
**Posts:**3041**Joined:**January 27th, 2019, 5:54 pm**Location:**A house, or perhaps the OCA board. Or [click to not expand]-
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### Re: Largest total computable function competition

Trying to understand BEAF.

the rules of beaf are as follows, right?

(# represents anything in the array)

the rules of beaf are as follows, right?

(# represents anything in the array)

Code: Select all

```
{a,1,#} = a
{a,b} = a^b
{#,1} = {#}
{a,b,1,1,1...1,d,#} (any from 1 1 in-between to many many many ones in-between) = {a,b,a,a,a...,{a,b-1,1,1,1...1,d,#},d-1,#}
{a,b,d,#} = {a,{a,b-1,d,#},d-1,#}
```

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

- testitemqlstudop
**Posts:**1323**Joined:**July 21st, 2016, 11:45 am**Location:**in catagolue-
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### Re: Largest total computable function competition

Yes, that's right for unidimensional arrays.

But I don't get anything above 2d

But I don't get anything above 2d

- BlinkerSpawn
**Posts:**1938**Joined:**November 8th, 2014, 8:48 pm**Location:**Getting a snacker from R-Bee's

### Re: Largest total computable function competition

Let the first non-one entry after b be the "active point" P and the separator before it S.testitemqlstudop wrote: ↑October 4th, 2019, 9:30 pmYes, that's right for unidimensional arrays.

But I don't get anything above 2d

Decrease S according to the following rules, and call the result S':

Code: Select all

```
{k #} -> {k-1 #}
{# 1,k #} -> {# b,k-1}
If this separator has another separator at its active point, reduce it the same way.
```

Examples:

Code: Select all

```
2,3,1{1}2 -> 2,3,1,1,1,2
2,3,1{3}3{4}2 -> 2,3,1{2}1{2}1{2}2{3}2{4}2
2,3,1{1,3}2 -> 2,3,1{3,2}1{3,2}1{3,2}2
2,3,1{1{3}2}2 -> 2,3,1{1{2}1{2}1{2}2}2
```

- testitemqlstudop
**Posts:**1323**Joined:**July 21st, 2016, 11:45 am**Location:**in catagolue-
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### Re: Largest total computable function competition

FIrst order set theory is much stronger, and you might as well take the maximum running time:

n ↦ The maximum running time of a Turing machine (with no input) such that there exists a proof of halting in first order set theory (under the domain of the vN universe) using fewer than n symbols.

I think this is some weak version of the Rayo function now.

### Re: Largest total computable function competition

Right. But the problem with your function is that you can't prove that it's total using ZF(C), which is the conventional axiom set for doing mathematics. I chose second order PA because it was the strongest axiom set I could think of that ZF can prove sound. Otherwise we could grow even faster by using ZF + large cardinal axiom.testitemqlstudop wrote: ↑October 4th, 2019, 11:11 pmFIrst order set theory is much stronger, and you might as well take the maximum running time:

n ↦ The maximum running time of a Turing machine (with no input) such that there exists a proof of halting in first order set theory (under the domain of the vN universe) using fewer than n symbols.

I think this is some weak version of the Rayo function now.

- testitemqlstudop
**Posts:**1323**Joined:**July 21st, 2016, 11:45 am**Location:**in catagolue-
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### Re: Largest total computable function competition

Wait, ZFC can't prove itself?