Can you give some examples? I still don't really get it.Moosey wrote:The Veblen function is actually pretty simple--

phi_0(n) = w^n,

phi_n+1(0) = the 0th fixed point of phi_n(a),

phi_n(0) for limit ordinal n = the limit of {phi_(n[0])(0), phi_(n[1]), (0) phi_(n[2])(0),...},

phi_n(m+1) = the next fixed point like phi_n(m),

phi_n(m) for limit ordinal m = the limit of {phi_n(m[0]),phi_n(m[1]),phi_n(m[2]),...}.

The multivariable Veblen function is basically the same, but it's organized like so: phi(mth entry, m-1th entry,...3rd entry,2nd entry, 1st entry)

The (n+1)th entry indicates fixed points in the nth entry, so gamma_0 = phi(1,0,0) because it's a|->phi_a(0)

## Ordinals in googology

- gameoflifemaniac
**Posts:**852**Joined:**January 22nd, 2017, 11:17 am**Location:**There too

### Re: Ordinals in googology

https://www.youtube.com/watch?v=q6EoRBvdVPQ

One big dirty Oro. Yeeeeeeeeee...

One big dirty Oro. Yeeeeeeeeee...

- Moosey
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### Re: Ordinals in googology

phi_1(0) = e_0gameoflifemaniac wrote: ↑November 2nd, 2019, 10:15 amCan you give some examples? I still don't really get it.Moosey wrote:The Veblen function is actually pretty simple--

phi_0(n) = w^n,

phi_n+1(0) = the 0th fixed point of phi_n(a),

phi_n(0) for limit ordinal n = the limit of {phi_(n[0])(0), phi_(n[1]), (0) phi_(n[2])(0),...},

phi_n(m+1) = the next fixed point like phi_n(m),

phi_n(m) for limit ordinal m = the limit of {phi_n(m[0]),phi_n(m[1]),phi_n(m[2]),...}.

The multivariable Veblen function is basically the same, but it's organized like so: phi(mth entry, m-1th entry,...3rd entry,2nd entry, 1st entry)

The (n+1)th entry indicates fixed points in the nth entry, so gamma_0 = phi(1,0,0) because it's a|->phi_a(0)

phi_2(5) = z_5

phi_4(0) = a|->h_a

phi_w(0) = sup{e_0,z_0,h_0,...}

phi_w+1(0) = a|->phi_w(a)

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?"

- gameoflifemaniac
**Posts:**852**Joined:**January 22nd, 2017, 11:17 am**Location:**There too

### Re: Ordinals in googology

SoMoosey wrote: ↑November 2nd, 2019, 10:34 amphi_1(0) = e_0gameoflifemaniac wrote: ↑November 2nd, 2019, 10:15 amCan you give some examples? I still don't really get it.Moosey wrote:The Veblen function is actually pretty simple--

phi_0(n) = w^n,

phi_n+1(0) = the 0th fixed point of phi_n(a),

phi_n(0) for limit ordinal n = the limit of {phi_(n[0])(0), phi_(n[1]), (0) phi_(n[2])(0),...},

phi_n(m+1) = the next fixed point like phi_n(m),

phi_n(m) for limit ordinal m = the limit of {phi_n(m[0]),phi_n(m[1]),phi_n(m[2]),...}.

The multivariable Veblen function is basically the same, but it's organized like so: phi(mth entry, m-1th entry,...3rd entry,2nd entry, 1st entry)

The (n+1)th entry indicates fixed points in the nth entry, so gamma_0 = phi(1,0,0) because it's a|->phi_a(0)

phi_2(5) = z_5

phi_4(0) = a|->h_a

phi_w(0) = sup{e_0,z_0,h_0,...}

phi_w+1(0) = a|->phi_w(a)

phi_0(0) = w

phi_0(1) = w^w

phi_0(2) = w^w^w

phi_1(0) = e0

phi_1(1) = e1

phi_1(2) = e2

phi_2(0) = z0

phi_2(1) = z1

phi_2(2) = z2

.

.

.

and

phi_w(0) = Gamma(0), kind of a limit of new kinds of fixed points?

Also, is

z1 = z0^z0^z0^z0^z0^...

z2 = z1^z1^z1^z1^z1^...

.

.

.

?

And how do you write

phi_3(n) = ?

phi_4(n) = ?

.

.

.

https://www.youtube.com/watch?v=q6EoRBvdVPQ

One big dirty Oro. Yeeeeeeeeee...

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- Moosey
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### Re: Ordinals in googology

No, phi_0(0) = 1, phi_0(1) = w, phi_0(2) = w^2gameoflifemaniac wrote: ↑November 2nd, 2019, 11:19 amSoMoosey wrote: ↑November 2nd, 2019, 10:34 amphi_1(0) = e_0gameoflifemaniac wrote: ↑November 2nd, 2019, 10:15 am

Can you give some examples? I still don't really get it.

phi_2(5) = z_5

phi_4(0) = a|->h_a

phi_w(0) = sup{e_0,z_0,h_0,...}

phi_w+1(0) = a|->phi_w(a)

phi_0(0) = w

phi_0(1) = w^w

phi_0(2) = w^w^w

Yesgameoflifemaniac wrote: ↑November 2nd, 2019, 10:15 amphi_1(0) = e0

phi_1(1) = e1

phi_1(2) = e2

phi_2(0) = z0

phi_2(1) = z1

phi_2(2) = z2

No. gamma_0 = phi_gamma_0(0). phi_w(0) << gamma_0. Actually I don't know why phi_w(0), which is a rather important ordinal being the supremum of {1,e_0, z_0, ...} does not have a name besides phi_w(0), but that's a real problem because everyone (myself included) that knows about ordinals has (or is) mistakenly thought that gamma_0 was the supremum of {1,e_0,z_0}...gameoflifemaniac wrote: ↑November 2nd, 2019, 10:15 am.

.

.

and

phi_w(0) = Gamma(0), kind of a limit of new kinds of fixed points?

I think those are e_(z_0 +1) and e_(z_1 +1), not the much larger z_1 and z_2.gameoflifemaniac wrote: ↑November 2nd, 2019, 10:15 amAlso, is

z1 = z0^z0^z0^z0^z0^...

z2 = z1^z1^z1^z1^z1^...

.

.

.

?

I think that for limit ordinal n > some point between e_0 and z_0, e_(n+1) = the 0th a|->n^a, or n^^w, which is why e_(W+1) = the 0th a|->W^a

the former is η_n, also known as the nth a|->z_a. η_0 can be though of as z_z_z_z_z_z_z...

The latter is the nth a|->η_a.

Generally we call things beyond eta-naught by their names in the Veblen function anyways, so there isn't a much better (or any other canonical) name for phi_4(n) than phi_4(n).

Last edited by Moosey on November 2nd, 2019, 11:39 am, edited 1 time in total.

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?"

- gameoflifemaniac
**Posts:**852**Joined:**January 22nd, 2017, 11:17 am**Location:**There too

### Re: Ordinals in googology

Ok, now I understand everything, but this

how do you construct e1, e2, e3... from e0; z1, z2, z3... from z0; and η1, η2, η3... from η0?

And how do you define Gamma(1), Gamma(2), Gamma(3)...?

What's the ϑ function, θ function, ψ function and Ω? And can you explain how does the multivariable phi function work?

how do you construct e1, e2, e3... from e0; z1, z2, z3... from z0; and η1, η2, η3... from η0?

And how do you define Gamma(1), Gamma(2), Gamma(3)...?

What's the ϑ function, θ function, ψ function and Ω? And can you explain how does the multivariable phi function work?

https://www.youtube.com/watch?v=q6EoRBvdVPQ

One big dirty Oro. Yeeeeeeeeee...

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- testitemqlstudop
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### Re: Ordinals in googology

Is the well ordering of w_1 provable?

- Moosey
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### Re: Ordinals in googology

e_1 = sup{e_0 +1,w^(e_0 +1), w^w^(e_0 +1),...}gameoflifemaniac wrote: ↑November 2nd, 2019, 11:38 amOk, now I understand everything, but this

how do you construct e1, e2, e3... from e0; z1, z2, z3... from z0; and η1, η2, η3... from η0?

e_2 = sup{e_1 +1,w^(e_1 +1), w^w^(e_1 +1),...}

...

z_1 = sup{z_0 +1,e_(z_0 +1),e_e_(z_0 +1)...}

...

h_1 = sup{h_0 +1,z_(h_0 +1),z_z_(h_0 +1)...}

For the epsilons, you could alternatively say that, very loosely,

e_(n+1) = (e_n)^^w

The next fixed points of phi_a(0) after gamma_0gameoflifemaniac wrote: ↑November 2nd, 2019, 11:38 amAnd how do you define Gamma(1), Gamma(2), Gamma(3)...?

Those are OCFs, which have gotten a bit of discussion here. Ω is w_1 = the first uncountable ordinalgameoflifemaniac wrote: ↑November 2nd, 2019, 11:38 amWhat's the ϑ function, θ function, ψ function and Ω?

I already did:gameoflifemaniac wrote: ↑November 2nd, 2019, 11:38 amAnd can you explain how does the multivariable phi function work?

So the 3rd entry does to the second entry what the second entry does to the first--series of fixed points for the previous entry

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?"

- gameoflifemaniac
**Posts:**852**Joined:**January 22nd, 2017, 11:17 am**Location:**There too

### Re: Ordinals in googology

Thanks now I understand everything. But what about w-1? Does it exist?

https://www.youtube.com/watch?v=q6EoRBvdVPQ

One big dirty Oro. Yeeeeeeeeee...

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- toroidalet
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### Re: Ordinals in googology

Isn't well-orderedness part of the definition of an ordinal?

If I'm missing something, then see this proof (set X to be the set of natural numbers to get the conclusion that ω_1 exists and is well-ordered)

If such an ordinal existed, it would be an infinite ordinal less than ω, which contradicts its definition as the smallest infinite ordinal. Limit ordinals are not successors and so subtraction is undefined in general.gameoflifemaniac wrote: ↑November 3rd, 2019, 12:33 pmThanks now I understand everything. But what about w-1? Does it exist?

However, surreal numbers include all ordinals and real numbers and things like ω-1, 1/ω, and √ω.

"Build a man a fire and he'll be warm for a day. Set a man on fire and he'll be warm for the rest of his life."

-Terry Pratchett

-Terry Pratchett

- gameoflifemaniac
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### Re: Ordinals in googology

Interesting. If Gamma(0) is phi_phi_phi...(0)(0)(0), is Gamma(1) phi_phi_phi...(1)(1)(1), Gamma(2) phi_phi_phi...(2)(2)(2) etc.?

Also, I heard somewhere that w+2 and 2+w aren't the same. Is that true?

Also, I heard somewhere that w+2 and 2+w aren't the same. Is that true?

https://www.youtube.com/watch?v=q6EoRBvdVPQ

One big dirty Oro. Yeeeeeeeeee...

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### Re: Ordinals in googology

no. Probably gamma_1 = sup{gamma_0+1,phi_gamma_0+1(0),phi_phi_gamma_0+1(0)(0),etc.}gameoflifemaniac wrote: ↑November 4th, 2019, 9:11 amInteresting. If Gamma(0) is phi_phi_phi...(0)(0)(0), is Gamma(1) phi_phi_phi...(1)(1)(1), Gamma(2) phi_phi_phi...(2)(2)(2) etc.?

HOW DID YOU MAKE IT THIS FAR WITHOUT KNOWING THAT WAS TRUE?gameoflifemaniac wrote: ↑November 4th, 2019, 9:11 amAlso, I heard somewhere that w+2 and 2+w aren't the same. Is that true?

Yes, w+2 = {0,1,2,3...w,w+1}, which is > 2+w = {0,0,0,1,2,3,...}

The same goes for w+1 and 1+w, or 2w and w2 (the former is = w, and the latter is larger)

On an unrelated note, I should update the definition of // from ah:

Code: Select all

```
ah_n{$_0//(b+1),$_1} = ah_n((($_0)(@^_n)ah_n{$_0//b})#{//b,$_1})
ah_n{$_0//0} = ah_n{$_0}
ah_n{$_0//b,$_1} = ah_n{$_0//(b[n]),$_1} for lim ord b
Else: apply ah's rules. (e.g. ah_{$_0//,0,0,w+1} = ah_{$_0//,w+1,w+1,w})
Where ($_1)(@^_n)a =
($_1)@(ah_n(($_1)(@^_n)(a-1))), a > 1,
ah_n{$_1}, a=1
```

Additionally, it allows for a better @@:

Code: Select all

```
ah_n{$_0@@0} = ah_n{$_0}
ah_n{$_0@@a,$_1} = ah_n{$_0@@a[n],$_1},a a lim ord
ah_n{$_0@@(a+1),$_1} = ah_n({$_0//}#{$_0@@(a),$_1}
Apply basic ah rules otherwise
```

Code: Select all

```
ah_n{$_0(@@b)0} = ah_n{$_0}
ah_n{$_0((@@b)a,$_1} = ah_n{$_0(@@b)a[n],$_1},a a lim ord
ah_n{$_0(@@0)(a+1),$_1} = ah_n({$_0//}#{$_0@@(a),$_1}
ah_n{$_0(@@b+1)(a+1),$_1} = ah_n({$_0(@@b)}#{$_0@@(a),$_1}
ah_n{$_0(@@b)$_2} = ah_n{$_0(@@b[n])$_2}, if b is a lim ord
Apply basic ah rules otherwise
```

ah_n{gamma_0(@@gamma_0)gamma_0}

Last edited by Moosey on November 5th, 2019, 7:26 am, edited 1 time in total.

My CA rules can be found here

Also, the tree game

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- gameoflifemaniac
**Posts:**852**Joined:**January 22nd, 2017, 11:17 am**Location:**There too

### Re: Ordinals in googology

Umm what do you mean? I knew that, but I wanted to be 100% sure.

Probably?

https://www.youtube.com/watch?v=q6EoRBvdVPQ

One big dirty Oro. Yeeeeeeeeee...

One big dirty Oro. Yeeeeeeeeee...

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

### Re: Ordinals in googology

Yes. Remember, phi(1,0,a) enumerates fixed points of phi(0,a,0).

Then phi(1,1,a) enumerates fixed points of phi(1,0,a), phi(1,2,a) the fixed points of phi(1,1,a), and so on until

phi(2,0,a) enumerates the fixed points of (1,a,0), phi(3,0,a) enumerates the fixed points of phi(2,a,0), and finally

phi(1,0,0,a) enumerates the fixed points of phi(a,0,0).

This continues until you reach the Small Veblen Ordinal, psi(W^W^w).

- testitemqlstudop
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### Re: Ordinals in googology

Several things:toroidalet wrote: ↑November 3rd, 2019, 3:43 pmIsn't well-orderedness part of the definition of an ordinal?

If I'm missing something, then see this proof (set X to be the set of natural numbers to get the conclusion that ω_1 exists and is well-ordered)

The well ordering of w_1 implies the well ordering of all countable ordinals.

Am I dumb or does that immediately prove the infinite-time computability of w_1^CK?

All proof systems have PTO < w_1, so wouldn't a proof be part of the most powerful proof system?

- gameoflifemaniac
**Posts:**852**Joined:**January 22nd, 2017, 11:17 am**Location:**There too

### Re: Ordinals in googology

Can you give me some examples of values of Weiermann's ϑ? Maybe I'll understand it this way

https://www.youtube.com/watch?v=q6EoRBvdVPQ

One big dirty Oro. Yeeeeeeeeee...

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- Moosey
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### Re: Ordinals in googology

Why that one? I've never even heard of that one! Maybe I can explain some simple OCF like Madore's ψ or some simpler one.gameoflifemaniac wrote: ↑November 5th, 2019, 1:56 pmCan you give me some examples of values of Weiermann's ϑ? Maybe I'll understand it this way

(that I made up)

Here's a simple example:

Code: Select all

```
C_0(α)={0,1,Ω}
C_n+1(α)={γ+δ,ψ(η)|γ,δ,η∈C_n(α);η<α}
C(α)=⋃(n<ω)C_n(α)
ψ(α)=min{β∈Ω|β∉C(α)}
```

ψ(1) = w^2 (The minimum value you cannot get by adding finitely many ws together)

ψ(n) = w^(n+1) for finite n

ψ(w) = w^w

etc.

ψ(e_0) = e_0

ψ(e_0 + 1) = e_0

... epsilon_0 for a long time

ψ(W) = e_0

ψ(W+1) = (e_0)w (I think)

The limit of ψ = e_w

Suppose we introduce multiplication:

Code: Select all

```
C_0(α)={0,1,Ω}
C_n+1(α)={γ+δ,γδ,ψ(η)|γ,δ,η∈C_n(α);η<α}
C(α)=⋃(n<ω)C_n(α)
ψ(α)=min{β∈Ω|β∉C(α)}
```

Still w

ψ(1) = the smallest ordinal you can't get by multiplying 0,1, or w finitely many times:

w^w

ψ(n) for finite n = w^^(n+1)

ψ(w) = e_0

ψ(w2) = I think e_1

ψ(wn) for finite n = I think e_(n-1)

ψ(w^2) = e_w in that case

...

Adding exponentiation would make it, for all purposes, madore's psi:

Code: Select all

```
That
C_0(α)={0,1,Ω}
C_n+1(α)={γ+δ,γδ,γ^δ,ψ(η)|γ,δ,η∈C_n(α);η<α}
C(α)=⋃(n<ω)C_n(α)
ψ(α)=min{β∈Ω|β∉C(α)}
```

Code: Select all

```
Madore's psi
C_0(α)={0,1,w,Ω}
C_n+1(α)={γ+δ,γδ,γ^δ,ψ(η)|γ,δ,η∈C_n(α);η<α}
C(α)=⋃(n<ω)C_n(α)
ψ(α)=min{β∈Ω|β∉C(α)}
```

If you want to see someone analyse it go to that link on the words Madore's psi.

Unfortunately you can't just extend it thru the hyper operators:

Code: Select all

```
C_0(α)={0,1,w,Ω}
C_n+1(α)={γ+δ,γδ,γ^δ,γ{ξ}δ,ψ(η)|γ,δ,ξ,η∈C_n(α);η<α}
C(α)=⋃(n<ω)C_n(α)
ψ(α)=min{β∈Ω|β∉C(α)}
```

But you can define this extremely simple one:

Code: Select all

```
C_0(α)={0,Ω}
C_n+1(α)={γ+δ,γδ,γ^δ,φ_γ(δ),ψ(η)|γ,δ,η∈C_n(α);η<α}
C(α)=⋃(n<ω)C_n(α)
ψ(α)=min{β∈Ω|β∉C(α)}
```

*zero*. Without so much as a 1 in the starter set.

With the finitely-many-entry Veblen function instead of the Veblen function, you could get to the SVO at 0, and with the any-amount-of-entries-in-the-function-including-infinitely-many you could get to the LVO.

Probably another good system would be

Code: Select all

```
C_0(α)={0,Ω}
C_n+1(α)={γ+δ,γδ,γ^δ,φ_γ(δ),ψ_0(η),w_γ|γ,δ,η∈C_n(α);η<α}
C(α)=⋃(n<ω)C_n(α)
ψ_0(α)=min{β∈Ω|β∉C(α)}
ψ_1(α)=sup(C(α))
```

In LaTeX:

Code: Select all

```
\begin{eqnarray*}
C_0(\alpha) &=& \{0, \Omega\}\\
C_{n+1}(\alpha) &=& \{\gamma + \delta, \gamma\delta,\gamma^\delta,\phi_{\gamma}(\delta),\omega_\gamma,\psi_0(\eta) | \gamma, \delta, \eta \in C_n (\alpha); \eta < \alpha\} \\
C(\alpha) &=& \bigcup_{n < \omega} C_n (\alpha) \\
\psi_0(\alpha) &=& \min\{\beta \in \Omega|\beta \notin C(\alpha)\} \\
\psi_1(\alpha) &=& \sup(C(\alpha)) \\
\end{eqnarray*}
```

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?"

- gameoflifemaniac
**Posts:**852**Joined:**January 22nd, 2017, 11:17 am**Location:**There too

### Re: Ordinals in googology

Because it's used to write any countale ordinal larger than Gamma(0) and smaller than w_1CK in the googology wiki.

Ackermann's ordinal is Weiermann's ϑ(Ω^3), for example, but what does that mean?

https://www.youtube.com/watch?v=q6EoRBvdVPQ

One big dirty Oro. Yeeeeeeeeee...

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- Moosey
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**Contact:**

### Re: Ordinals in googology

If you need that one specifically...gameoflifemaniac wrote: ↑November 6th, 2019, 2:57 amBecause it's used to write any countale ordinal larger than Gamma(0) and smaller than w_1CK in the googology wiki.

Ackermann's ordinal is Weiermann's ϑ(Ω^3), for example, but what does that mean?

... Then they have a nice explanation on the wiki:

Essentially, it wrote: C(a,b) is the set of all ordinals constructible using only the following:

0, all ordinals less than b, and Ω.

Finite applications of c+d, κ↦ωκ [or w^c in other terms], and ϑ(n) for n < a, assuming n, c, and d are produceable using those methods or are 0 or W.

Then, ϑ(a) is the smallest ordinal b such that α∈C(a,b), and b is greater than all the countable ordinals in C(a,b)

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?"

- gameoflifemaniac
**Posts:**852**Joined:**January 22nd, 2017, 11:17 am**Location:**There too

### Re: Ordinals in googology

I don't really understand the definition. I would get it if you'd gave me some examples.Moosey wrote: ↑November 6th, 2019, 7:39 amIf you need that one specifically...gameoflifemaniac wrote: ↑November 6th, 2019, 2:57 amBecause it's used to write any countale ordinal larger than Gamma(0) and smaller than w_1CK in the googology wiki.

Ackermann's ordinal is Weiermann's ϑ(Ω^3), for example, but what does that mean?

... Then they have a nice explanation on the wiki:Essentially, it wrote: C(a,b) is the set of all ordinals constructible using only the following:

0, all ordinals less than b, and Ω.

Finite applications of c+d, κ↦ωκ [or w^c in other terms], and ϑ(n) for n < a, assuming n, c, and d are produceable using those methods or are 0 or W.

Then, ϑ(a) is the smallest ordinal b such that α∈C(a,b), and b is greater than all the countable ordinals in C(a,b)

Also, am I bothering you in this topic? You all are asking much more complicated things than me and I don't want to disturb you.

https://www.youtube.com/watch?v=q6EoRBvdVPQ

One big dirty Oro. Yeeeeeeeeee...

One big dirty Oro. Yeeeeeeeeee...

- Moosey
**Posts:**3044**Joined:**January 27th, 2019, 5:54 pm**Location:**A house, or perhaps the OCA board. Or [click to not expand]-
**Contact:**

### Re: Ordinals in googology

I'm not that good with more complicated OCFs, so I can't answer that question.gameoflifemaniac wrote: ↑November 6th, 2019, 1:16 pmI don't really understand the definition. I would get it if you'd gave me some examples.

No. Personally I like the activity. It keeps this thread up and I desperately want a response to my ah stuff (like this):gameoflifemaniac wrote: ↑November 6th, 2019, 1:16 pmAlso, am I bothering you in this topic? You all are asking much more complicated things than me and I don't want to disturb you.

Moosey wrote: ↑November 4th, 2019, 9:54 amOn an unrelated note, I should update the definition of // from ah:

This makes expressions such as ah_5{w+5//7,w+100223} valid.Code: Select all

`ah_n{$_0//(b+1),$_1} = ah_n((($_0)(@^_n)ah_n{$_0//b})#{//b,$_1}) ah_n{$_0//0} = ah_n{$_0} ah_n{$_0//b,$_1} = ah_n{$_0//(b[n]),$_1} for lim ord b Else: apply ah's rules. (e.g. ah_{$_0//,0,0,w+1} = ah_{$_0//,w+1,w+1,w}) Where ($_1)(@^_n)a = ($_1)@(ah_n(($_1)(@^_n)(a-1))), a > 1, ah_n{$_1}, a=1`

Additionally, it allows for a better @@:This, of course, can be extended indefinitely:Code: Select all

`ah_n{$_0@@0} = ah_n{$_0} ah_n{$_0@@a,$_1} = ah_n{$_0@@a[n],$_1},a a lim ord ah_n{$_0@@(a+1),$_1} = ah_n({$_0//}#{$_0@@(a),$_1} Apply basic ah rules otherwise`

Which of course leads to my new fastest-growing-function (by Moosey):Code: Select all

`ah_n{$_0(@@b)0} = ah_n{$_0} ah_n{$_0((@@b)a,$_1} = ah_n{$_0(@@b)a[n],$_1},a a lim ord ah_n{$_0(@@0)(a+1),$_1} = ah_n({$_0//}#{$_0@@(a),$_1} ah_n{$_0(@@b+1)(a+1),$_1} = ah_n({$_0(@@b)}#{$_0@@(a),$_1} ah_n{$_0(@@b)$_2} = ah_n{$_0(@@b[n])$_2}, if b is a lim ord Apply basic ah rules otherwise`

ah_n{gamma_0(@@gamma_0)gamma_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?"

- gameoflifemaniac
**Posts:**852**Joined:**January 22nd, 2017, 11:17 am**Location:**There too

### Re: Ordinals in googology

What is the large Veblen ordinal? I know it's defined by α↦φ_Ω^α(0), but can you even talk about phi of an uncountable cardinal?

What is phi(1,1,0), phi(2,0,0)?

And is the Ackermann ordinal equal to Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ...?

What is phi(1,1,0), phi(2,0,0)?

And is the Ackermann ordinal equal to Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ...?

Last edited by gameoflifemaniac on November 8th, 2019, 4:23 pm, edited 1 time in total.

https://www.youtube.com/watch?v=q6EoRBvdVPQ

One big dirty Oro. Yeeeeeeeeee...

One big dirty Oro. Yeeeeeeeeee...

- Moosey
**Posts:**3044**Joined:**January 27th, 2019, 5:54 pm**Location:**A house, or perhaps the OCA board. Or [click to not expand]-
**Contact:**

### Re: Ordinals in googology

No, itś defined like this:gameoflifemaniac wrote: ↑November 8th, 2019, 11:42 amWhat is the large Veblen ordinal? I know it's defined by α↦φ_Ω^α(0), but can you even talk about phi of an uncountable cardinal?

Code: Select all

```
1. w
2. phi(1,0,0,0,0,0,0,0,...) w/ w 0´s (SVO)
3. phi(1,0,0,0,0,0,...) w/ that many 0s
4. phi(1,0,0,0,...) w/ that many 0s
...
The LVO is the supremum of all of these
```

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?"

- gameoflifemaniac
**Posts:**852**Joined:**January 22nd, 2017, 11:17 am**Location:**There too

### Re: Ordinals in googology

Thisgameoflifemaniac wrote: ↑November 8th, 2019, 11:42 amWhat is phi(1,1,0), phi(2,0,0)?

And is the Ackermann ordinal equal to Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ...?

Also, I can't appreciate your ah function cuz small brain

https://www.youtube.com/watch?v=q6EoRBvdVPQ

One big dirty Oro. Yeeeeeeeeee...

One big dirty Oro. Yeeeeeeeeee...

- Moosey
**Posts:**3044**Joined:**January 27th, 2019, 5:54 pm**Location:**A house, or perhaps the OCA board. Or [click to not expand]-
**Contact:**

### Re: Ordinals in googology

And is the Ackermann ordinal equal to Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ...?gameoflifemaniac wrote: ↑November 8th, 2019, 11:42 amWhat is phi(1,1,0), phi(2,0,0)?

And is the Ackermann ordinal equal to Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ_Γ...?

No.

I think min a|->gamma_a = phi(1,1,0) but I could be wrong.

phi(2,0,0) = min a|->phi(1,a,0) much like how gamma_0 = phi(1,0,0) = min a|->phi(a,0)

Simply put, ah is my extension of something by pkmnQgameoflifemaniac wrote: ↑November 8th, 2019, 3:57 pmAlso, I can't appreciate your ah function cuz small brain

Since Ę_0,a(n) is hardly more powerful than f_a(n), I extended it much farther-- for instance, ah_n(a@w) is at least f_a*n-ish, and ah_n(a(@@a)a) is probably much more powerful for large enough a. Using the extension of ah rules I will write now.

(Can someone check how powerful it is? Is it e_(a+1)-ish?)

ah rules:

Code: Select all

```
Part one:
define $_n as any entries (including no entries) in an array.
It’s my symbol for we-don’t-care entries.
In any one use of any one rule, if n is the same, $_n is the same
a#b = concatenation of a and b
n@m =
n, m = 1
{n}#(n@(m-1)), m > 1, m not a lim ord
n@(m[n]) if m is a lim ord
g(a,n,B) =
ah_g(a-1,n,B) {B}, a > 1;
n, a = 0
ah definition:
Rule 1.
ah^a_n {$_0} = g(a,n,$_0)
Rule 2.
ah_n ({$_1}#{z}) = ah_n {$_1}, z = 0
Rule 3.
ah_n{} = n+1
Rule 4.
ah_n{a+1,$_2} = ah^n_n{a,$_2}
Rule 5.
ah_n{a,$_3} = ah_n{a[n],$_3}, a a lim ord
Rule 6.
ah_n ((0@b)#{a+1,$_4}) = ah_(n+1) (((a+1)@b)#{a,$_4}), b > 0
Rule 7.
ah_n ((0@b)#{a,$_5}) = ah_n ((a@b)#{a[n],$_5}), a a lim ord & b > 0
The rest of the array rules:
ah_n{$_0//(b+1),$_1} = ah_n((($_0)(@^_n)ah_n{$_0//b})#{//b,$_1})
ah_n{$_0//0} = ah_n{$_0}
ah_n{$_0//b,$_1} = ah_n{$_0//(b[n]),$_1} for lim ord b
Else: apply ah's 7 rules, starting after legion bar. (e.g. ah_{$_0//,0,0,w+1} = ah_{$_0//,w+1,w+1,w})
($_1)(@^_n)a =
($_1)@(ah_n(($_1)(@^_n)(a-1))), a > 1,
ah_n{$_1}, a=1
ah_n{$_0@@0} = ah_n{$_0}
ah_n{$_0@@a,$_1} = ah_n{$_0@@a[n],$_1},a a lim ord
ah_n{$_0@@(a+1),$_1} = ah_n({$_0//}#{$_0@@(a),$_1}
Apply basic ah rules otherwise to everything after the @@
ah_n{$_0(@@b,$_2)0} = ah_n{$_0}
ah_n{$_0((@@b,$_2)a,$_1} = ah_n{$_0(@@b)a[n],$_1},a a lim ord
ah_n{$_0(@@0)(a+1),$_1} = ah_n({$_0//}#{$_0(@@0)(a),$_1}
ah_n{$_0(@@b+1,$_2)(a+1),$_1} = ah_n({$_0(@@b,$_2)}#{$_0@@(a),$_1}
ah_n{$_0(@@b,$_2)$_3} = ah_n{$_0(@@b[n],$_2)$_3}, if b is a lim ord
Apply basic ah rules otherwise to everything after the (@@$) or everything inside the (@@$) depending on what is necessary
These array rules can be applied without the ah_n of the array, but any n refers to the n in the ah subscript.
```

What is the power of ah_n{gamma_0(@@gamma_0)gamma_0} ?

EDIT:

Redefine legion bars and the like in a way that formalizes the array rules without the ah:

Code: Select all

```
if unspecified, n is the subscript in the first ah that is in front of the array.
Now we can redefine the rest of the array rules:
The rest of the array rules:
{$_0//(b+1),$_1} = ((($_0)(@^_n)ah_n{$_0//b})#{//b,$_1})
{$_0//0} = {$_0}
{$_0//b,$_1} = {$_0//(b[n]),$_1} for lim ord b
Else: apply ah's 7 rules, starting after legion bar. This includes nesting and incrementing the subscript (e.g. ah_n{$_0//,0,0,w+1} = ah_(n+1){$_0//,w+1,w+1,w})
($_1)(@^_n)a =
($_1)@(ah_n(($_1)(@^_n)(a-1))), a > 1,
ah_n{$_1}, a=1
{$_0@@0} = {$_0}
{$_0@@a,$_1} = {$_0@@a[n],$_1},a a lim ord
{$_0@@(a+1),$_1} = ({$_0//}#{$_0@@(a),$_1}
Apply basic ah rules otherwise to everything after the @@. See legion bar notes for more details
{$_0(@@b,$_2)0} = {$_0}
{$_0((@@b,$_2)a,$_1} = {$_0(@@b)a[n],$_1},a a lim ord
{$_0(@@0)(a+1),$_1} = ({$_0//}#{$_0(@@0)(a),$_1}
{$_0(@@b+1,$_2)(a+1),$_1} = ({$_0(@@b,$_2)}#{$_0@@(a),$_1}
{$_0(@@b,$_2)$_3} = {$_0(@@b[n],$_2)$_3}, if b is a lim ord
Apply basic ah rules otherwise to everything after the (@@$) or everything inside the (@@$) depending on what is necessary. See legion bar notes for more details.
(Now things such as ah_n{a(@@(a(@@a)a))a} are well defined.)
{$_0@@@(a+1),$_1} = {$_0(@@($_0@@@a,$_1))$_0,$_1}
{$_0@@@0} = {$_0}
{$_0@@@a,$_1} = {$_0@@@a[n],$_1}, a a lim ord.
```

Last edited by Moosey on November 16th, 2019, 8:28 am, edited 2 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?"

- gameoflifemaniac
**Posts:**852**Joined:**January 22nd, 2017, 11:17 am**Location:**There too

### Re: Ordinals in googology

Wow.

What is the Bachmann-Howard ordinal?

What is the Bachmann-Howard ordinal?

https://www.youtube.com/watch?v=q6EoRBvdVPQ

One big dirty Oro. Yeeeeeeeeee...

One big dirty Oro. Yeeeeeeeeee...