Showing that Graham Number $G_{64}$, $TREE(3)$, loader.c output $D^{5}(99)$, and $Rayo(10^{100})$ are listed in order of increasing size

Question

Let

1. $G_{64}$ be Graham's Number:

https://googology.wikia.org/wiki/Graham%27s_number

2. $TREE(3)$ be a particular value of a special sequence $TREE(k)$:

https://googology.wikia.org/wiki/TREE_sequence

3. $D^{5}(99)$ be an output of the loader.c program:

Googology - Loader's number

4. $Rayo(10^{100})$ be Rayo's number:

https://googology.wikia.org/wiki/Rayo%27s_number

How to prove that $1.<2.<3.<4.$ ?

First of all, I know that there were several related questions, but, in fact, nobody has given a proper proof of any of these inequalities. That's why I wrote this question.

Let's take a look at all cases.

$1.<2.$

It is said that the sequence that generates Graham numbers grows more slowly than the sequence $TREE(k)$. But here are my questions:

• Where are the proofs, sources of this assertion?
• Even if some function $f$ grows faster than $g$ it doesn't prove that $f(n)>g(n)$ for certain value $n$.

$2.<3.$

Here is my idea to prove this:

First make a program that describes (not calculates) $TREE(3)$. Start it and see how much time it takes before you see a message (for example, "Hello World") on the screen. The same with the code that describes loader.c. Compare these numbers. The number that has greater time is greater. Here are other problems.

• The code from the site I gave you in the link doesn't work on Code::Blocks.
• Is my reasoning even correct? If not (which is very likely), then how to do this?

$3.<4.$

There is only one thing to know: the amount of symbols expressing $D^{5}(99)$ in language of first-order set theory. If that number is smaller than $10^{100}$, then the proof follows.

• How we can show that?

Answer 1

In the article https://cs.nyu.edu/pipermail/fom/2006-March/010260.html

Friedman showed that $$\text{TREE(3)} >\text{tree}(n(4)) + n(4)\tag{1}$$

where $n()$ is Friedman's block-subsequence function and $\text{tree}(n)$ is defined as the length of a longest sequence of unlabelled trees $T_1,T_2,\ldots$, such that, for all $i$, $T_i$ has at most $n+i$ vertices, and for all $i,j$ with $i<j,$ $T_i$ is not homeomorphically embeddable into $T_j.$

(A sketch of his proof is appended to the present answer.)

Furthermore, at https://mathoverflow.net/a/95588/20307 there is a proof that

$$\text{tree}(n)\geq H_{\vartheta (\Omega^{\omega}, 0)}( n) - n\tag{2}$$

where $H_{\alpha}$ is an accelerated version of the Hardy hierarchy; hence, as pointed out there,

$$\text{TREE(3)} > H_{\vartheta (\Omega^{\omega}, 0)}(n(4)).$$

We note that the RHS is clearly larger than $G_{64},$ because $H_{\omega^{\omega+2}}(3)>f_{\omega+2}(3)>G_{64},$ where $f_\alpha$ is the usual fast-growing hierarchy.

---

The following is an attempt to briefly elucidate the proof that $$\text{TREE}(3) > \text{tree}(q) + q,$$ where $q>n(4)$, based on the tree sequence and symbol-encodings used by Friedman in the article cited above. I'll use balanced parentheses of types $(\,),[\,],\{\,\}$ to denote vertices labelled with $1,2,3$ respectively.

T_1   {}
T_2   [[]]
T_3   [()()]
T_4   [((()))]
T_5   ([][][][])
T_6   ([][][](()))
T_7   ([][](()()()))
T_8   ([][](()(())))
T_9   ([][](((((()))))))
T_10  ([][]((((())))))
T_11  ([][](((()))))
T_12  ([][]((())))
T_13  ([][](()))
T_14  ([][]())
T_15  (A(B(((([]))))))
      (A(B((([])))))
      (A(B(([]))))
      (A(B([])))
      (A(B[]))
T_20  (C(D(E(((([])))))))
      (C(D(E((([]))))))
      (C(D(E(([])))))
      (C(D(E([]))))
      (C(D(E[])))
T_25  (F(G(H(I(((([]))))))))
...
T_30  (J(K(L(M(N(((([])))))))))
...
T_q   (X(Y(...(Z((((([])))))...))))  where q = 10 + 5p
T_q+1 (((...(())...)))  with q+1 ()s
...
T_q+tree(q)  ()

Here, each letter A,B,C,... denotes one of the following symbol-codes for a 4-symbol alphabet {1,2,3,4}:

(((())))   <- codes the symbol 1
((()()))   <- codes the symbol 2
(()()())   <- codes the symbol 3
((())())   <- codes the symbol 4

Now, by Friedman's results on the function $n()$, there exists a $p$-long sequence of words $x_1,...,x_p$ on alphabet $\{1,2,3,4\}$ such that $|x_i| = i+1$ and for all $i<j$, $x_i$ is not a subsequence of $x_j$, where $p = {n(4)-1\over 2}$.

So tree-embeddings are avoided by choosing the symbol-codes $A,B,C,...$ such that $AB$ encodes $x_1$, $CDE$ encodes $x_2$, $FGHI$ encodes $x_3$, etc.

Hence the sequence of trees $T_1,...,T_q$ (where $q = 10+5p>n(4)$) is such that $|T_i| \le i$ and for all $i<j$, $T_i$ is not homeomorphically embeddable in $T_j.$

Then the sequence continues $T_{q+1},...,T_{q+\text{tree}(q)}$ for another $\text{tree}(q)$ trees after $T_q$, these trees being constructed with $(\,)$-vertices only. QED

Answer 2

1. < 2.

$TREE(3) > tree^{tree^{tree^{tree^{tree^8(7)}(7)}(7)}(7)}(7)$ from Math StackExchange answer "A question about relation between two fast growing sequences.", based on another answer "Why is TREE(3) so big? (Explanation for beginners)"

$tree^{tree^{tree^{tree^{tree^8(7)}(7)}(7)}(7)}(7) \geq tree(4)$ by monotonicity of the weak tree function.

$tree(4) > f_{\varepsilon_0}(G_{64})$ from Takayuki Kihara's blog post "Lower bounds for $\mathrm{tree}(4)$ and $\mathrm{tree}(5)$"

$f_{\varepsilon_0}(G_{64}) > G_{64}$ by the fact that under the Wainer hierarchy, $f_\alpha(n) > n$ for all ordinals $\alpha$

Thus, $TREE(3) > G_{64}$

For a more general proof that $G_n$ will never outgrow $TREE(n)$:

The following structure is always possible as a game in $TREE(n)$ for $n\geq3$: Initial move removes one label, $n^2-3n+2$ moves removing all possible pairings of labels using 2 node trees (leaving only trees consisting entirely of 1 label as possible moves), $tree(n^2-3n+3)$ moves exhausting one remaining label, $tree(tree(n^2-3n+3)+n^2-3n+3)$ moves exhausting the next label, etc. Thus a lower bound on $TREE(n)$ for $n\geq3$ is $tree^{n-1}(n^2-3n+3)$

$\forall n, f_{\omega+2}(n) \geq f_{\omega+1}(n) \approx G_n$ and equality is only obtained for $n=1$

More pertinently, $\forall n\geq2, f_{\omega+2}(n) > G_n$

$\forall n > 3, tree(n^2-3n+3) > f_{SVO}(n)$ (by Kihara above)

Now the fundamental sequence used for $f_{SVO}(n)$ in this comparison is, to my knowledge:

$\vartheta(\Omega^n) = \phi(1@n)\\ SVO = \vartheta(\Omega^\omega)\\ SVO[n] = \vartheta(\Omega^n)= \phi(1@n)\\$ And then standard fundamental sequences for the finitary Veblen function. A key observation that can be made from these sequences is that, for $n\geq2, f_{SVO}(n)$ will trace a path of normal functions that passes through $f_{\phi(1,0)}(n) = f_{\varepsilon_0}(n)$ at some point. Thus if we can show that $f_{\varepsilon_0}(n)$ is never outgrown by $G_n$, that is sufficient to prove that $TREE(n)$ is not either.

$\forall n\geq3, f_{\varepsilon_0}(n) \geq f_{\omega^3}(n) > f_{\omega^2}(n) > f_{\omega+2}(n) > G_n$

Thus: $\forall n>3, TREE(n) \geq tree^{n-1}(n^2-3n+3) \geq tree(n^2-3n+3) > f_{SVO}(n) \geq f_{\varepsilon_0}(n) > G_n$

The proof for $n=3$ is detailed above, and the values for $n\leq2$ can be computed directly.

2. < 3.

I am unaware of any existing proof of this, but I can envisage how it would be done. You would start by writing a program in the Calculus of Constructions (the lambda calculus system that loader.c diagonalises over) that computes $TREE(3)$, and find an upper bound on the number of bits such a program takes. Then prove that that bound is less than $D^4(99)$, and thus that the computation of $TREE(3)$ is a part of Loader's number (which is $D^5(99)$ and therefore Loader's number is at least as large as $TREE(3)$)

3. < 4.

$\text{Loader's Number} < BB(1015) < BB(2^{65536}-1) \leq Rayo(7339) < Rayo(10^{100})$

The proof that $\text{Loader's number} < BB(1015)$ is that "output_loader.txt", a $1015$-state, $2$-symbol Turing machine written by CatIsFluffy, computes (a slightly larger variant of) Loader's number steps before halting. For $BB(2^{65536}-1) \leq Rayo(7339)$, a proof that $BB(2^{65536}-1) \leq Rayo(7901)$ is on Emk's Googology Wiki post "A Rayo name larger than BB(10^100)", but this is not for $Rayo(7339)$.