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Simply Beautiful Art

01:38

(almost) Absolute slowest growing ordinal function in the history of everything:

$$C(\alpha)_0=\{n:n<\omega\}\\C(\alpha)_{n+1} =C(\alpha)_n\cup\{\psi_\gamma(\eta): \gamma,\eta\in C(\alpha)_n, \eta<\alpha\}\\C(\alpha) =\bigcup_{n<\omega}C(\alpha)_n\\\psi_\beta( \alpha)=\min\{\delta:\delta \notin C(\alpha),\delta\ge \omega_\beta\}$$

@Secret Even you can probably see what's wrong with this. (Nothing is actually wrong, but it doesn't amount to a very fast growing function)

For starters, $\psi_\beta(0)=\omega_\beta$.

And, well, $\psi_\beta(1)=\omega_\beta+1$ for most $\beta$...

Secret

01:54

It seems $\psi_0(0)$ is specified before C(0) is even made?

Simply Beautiful Art

@Secret Uh, no.

$C(0)=C(0)_0$, since there are no $\eta<0$ for us to use the $C(\alpha)_{n+1}$ step.

Thus, the minimum ordinal which is $\ge\omega_\beta$ is $\omega_\beta=\psi_\beta(0)$.

Secret

Ah i see

Hmm I will think about how to convert that to psi notation later, cause the recursive version is much easier to work with

Simply Beautiful Art

I am curious though as to what the value of $$\psi_0(\omega_{ \omega_{ \omega_{ \ddots}}})$$ is.

@Secret For this OCF, I think this version is much easier to work with. There's practically only one thing that occurs in this, and its very easy to calculate values.

11 hours later…

Leaky Nun

12:50

@Secret could you help me with something?

I'm trying to axiomatize the theory of ordinals

the language has signature {0,ω,S,+,⋅,<}

ordering:
1. (∀m)(¬(m<m))
2. (∀a)(∀b)(∀c)(a<b⟹b<c⟹a<c)
3. (∀a)(∀b)(a<b⟹¬(b<a))
4. (∀a)(∀b)(a<b∨b<a)
constants:
1. (∀m)(¬(m<0))
2. 0<ω
3. (∀m)(∃n)(m<ω⟹m=0∨m=S(n))
successor:
1. (∀a)(¬(S(a)=0))
2. (∀a)(¬(S(a)=ω))
3. (∀m)(∀n)(S(m)=S(n)⟹m=n)
4. (∀a)(a<S(a))
addition:
1. (∀m)(m+0=m)
2. (∀m)(∀n)(m+S(n)=S(m+n))
3. (∀m)(∀n)(∀y)((∀x)(x<n⟹(m+x)<y)⟹m+n<y)
4. (∀m)(∀n)(∀x)(m<n⟹x+m<x+n)
multiplication:
1. (∀m)(m⋅0=0)
2. (∀m)(∀n)(m⋅S(n)=(m⋅n)+m)
3. (∀m)(∀n)(∀y)((∀x)(x<n⟹(m⋅x)<y)⟹m⋅n<y)
transfinite induction:
1. (∀m)((∀n)(n<m⟹φ(n))⟹φ(m))⟹(∀m)φ(m)

Could you see if you can prove that 1+ω=ω?

Secret

hmm...

Leaky Nun

I should try feeding those to prover9

fix:
ordering 4 should read (∀a)(∀b)(a<b∨a=b∨b<a)

addition 3 should read (∀m)(∀n)(∀y)((∀x)(x<n⟹(m+x)<y)⟹m+n<y ∨ m+n=y)

multiplication 3 should read (∀m)(∀n)(∀y)((∀x)(x<n⟹(m⋅x)<y)⟹m⋅n<y ∨ m⋅n=y)

Leaky Nun

13:33

after exploring on prover9, addition 3 and multiplication 3 should have ¬(y=0) prepended somewhere

Secret

uh, y cannot be 0 anyway since nothing can be smaller than 0

(ω<ω+1⟹1+ω<y)⟹1+ω+1<y ∨ 1+ω+1=y

hmm...

does not seemed very helpful...

(1<ω⟹(1+1)<ω)⟹1+ω<ω ∨ 1+ω=ω

and using transfinite induction, you will end up proving 1<ω⟹(1+1)<ω⟹(1+1+1)<ω⟹(1+1+1+1)<ω⟹...

It is not clear if ω<ω will be resulted at all

(ω<ω⟹(1+ω)<ω)⟹1+ω<ω ∨ 1+ω=ω) is false premise, though

Secret

14:36

Q: Ordinal addition is not commutative

The simplest example is $$1+\omega=\omega \neq \omega +1$$ But I think it is also true that $$\omega ^\alpha+\omega^\beta=\omega ^\beta$$ for general ordinals $\alpha < \beta$ but I don't know how to prove this. In fact I don't even know how to prove $1+\omega^2=\omega^2$. I've tried induction...

ordinals

@LeakyNun It seems we cannot in general prove $\alpha < \beta \implies \alpha + \beta = \beta$ purely algebraically, unless we can include the notion of an order isomorphism somehow in the axioms

Leaky Nun

alright, thanks

Secret

I suspect, but have no guarentee that adding some right absorbing axiom whenever the right term is larger than the left term could reproduce all the nondecreasing behaviour of ordinals when operated on the left

e.g. (∀a)(∀b)(a<b⟹a+b<b)

7 hours later…

Leaky Nun

22:03

@Secret but that isn't true

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