Comparing big numbers. Let


*

*$G_{64}$ is a Graham Number:


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


*$TREE(3)$ is a particular value of a special sequence $TREE(k)$
https://googology.wikia.org/wiki/TREE_sequence


*$D^{5}(99)$ is a output of program loader.c


https://googology.wikia.org/wiki/Loader's_number


*$Rayo(googol)$ is a Rayo 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 question 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 sequence that generates Graham numbers grows slower than 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 code that describes (not calculate) $TREE(3)$ 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 another problems.


*

*The code from the site i gave you in the link doesn't work on codeblocks.

*Is my reasoning even correct? If not( wich is very likely) then how to do this?


$3.<4.$
There is only one thing to know: an amount of symbols expressing $D^{5}(99)$ in language of First-Order-Set-Theory .
If that number is smaller than $googol$ then proof follows.


*

*How we can show that?


Regards
 A: 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
