TITLE: The inconsistency of Graham Arithmetics plus $ \forall n, n < g_{64}$
QUESTION [12 upvotes]: As you all know, Ronald Graham just passed away. He is famous for many fabulous contributions to finite combinatorics, and much much  more, but perhaps none of them is as popular as the infamous Graham number (see here): $g_{64}$.
This number is truly huge, though in more recent times it has been dwarfed by several other "finite inaccessible numbers" (allow me to call them that way), such as Friedman's TREE(3) .
Anyway, it is big enough to play the role of "infinite number" in what follows.
Now, start with Robinson's arithmetics $Q$ and "beef it up" so that it contains all the recursive definitions necessary to talk about $g_{64}$ (for instance, all equations to define upper arrow notation).
As a tribute to Graham, let us call this theory  GRAHAM.
Define T as GRAHAM + $ \forall n, n < g_{64}$
Clearly, T is classically inconsistent. But: just like Parikh's feasible arithmetics, there are, as far as I see, no short proofs of its inconsistency without appealing to induction.
Now the question:

assuming a fixed proof system (say Gentzen), can we give a measure of
the shortest induction-free proof that  T is not consistent?

NOTE: I have mentioned Parikh's aritmetics, because obviously T and Parikh system share some common features.
But in his case, he keeps some induction and augments arithmetics with an additional unary predicate expressing feasibility for which induction does not apply , whereas here there is no such predicate, but on the other hand there is no induction whatsoever.

REPLY [7 votes]: The length of the least proof of contradiction in $\mathsf{Graham}+\forall n (n<g_{64})$ should be inbetween $(\log_2^*(g_{64}))^{1/N}$ and $(\ln^*(g_{64}))^{N}$, where $\ln^*(x)=\min\{n\mid \log_2^n(x)<0\}$ and $N$ is some reasonable small integer (which could be figured out by a careful examination of the proof).
So let me first spell out what exactly I take as $\mathsf{Graham}$. Namely I'll assume that $\mathsf{Graham}$ is $\mathsf{Q}$ with additional function $x\uparrow^{y}z$ and axioms

$x\uparrow^z0=x$
$x\uparrow^0S(y)=x(x\uparrow^0y)$
$x\uparrow^{S(z)}S(y)=x\uparrow^z(x\uparrow^{S(z)}y)$

Note that here $x\uparrow^{y}z$ should corresponds to $x\underbrace{\uparrow\ldots\uparrow}\limits_{y+1\;\text{arrows}}z+1$ rather than $x\underbrace{\uparrow\ldots\uparrow}\limits_{y\;\text{arrows}}z$. This change of notations is due to the fact that in $\mathsf{Q}$ we start naturals with $0$ rather than $1$.
Numbers $g_{n}$ thus would be denoted by the terms

$g_1=\underline{3}\uparrow^{\underline{3}}\underline{2}$
$g_{n+1}=\underline{3}\uparrow^{g_n}\underline{2}$

Proof of the upper bound essentially follows from Emil's answer. Using axioms of $\mathsf{Graham}$ by polynomial in $\log_2^*(g_{64})$ proof we would show that $g_{64}=\underline{3}\uparrow^{\underline{1}} \underline{b}$, for appropriate $b<\log_2^*(g_{64})$. Next,since $\uparrow^0$ is exponentiation operation we could apply Emil's analysis to show by a polynomial in $\log_2^*(g_{64})$ proof in $\mathsf{Graham}$ that $\lnot\underline{3}\uparrow^{\underline{1}} \underline{b}<\underline{3}\uparrow^{\underline{1}} \underline{b}$. This gives an upper bound for the length of the proof of contradiction.
I'll sketch how to give the lower bound using fulfillment sequences see [1].
I'll assume that we use the variant of first-order language with $\forall,\exists,\land,\lor$ and negations that could be only used with atomic formulas; for non-atomic $\varphi$ we denote as $\lnot\varphi$ the formula that is obtained from $\varphi$ by replacing $\forall$ with $\exists$, $\land$ with $\lor$, non-negated atomic $\psi$, with $\lnot \psi$ and vice versa.
Let $\tau=\langle \tau_i\mid i\le n\rangle$ be a sequence of non-decreasings subsets of some model $\mathfrak{M}_{\tau}$ of the language $L(\mathsf{Graham})$. We call $\tau$ of this form fulfulment sequences.  We define fulfilment relation $\tau, i\Vdash \varphi$, for $i\le n$ and $L(\mathsf{Graham})$ formulas $\varphi$ with parameters from $\tau_i$:

$\tau,i\Vdash \varphi\iff \mathfrak{M}\models \varphi$, if $\varphi$ is atomic or negation of atomic
$\tau,i\Vdash \varphi\land \psi \iff \tau,i\Vdash \varphi\text{ and }\tau,i\Vdash \psi$
$\tau,i\Vdash \varphi\lor \psi \iff \tau,i\Vdash \varphi\text{ or }\tau,i\Vdash \psi$
$\tau,i\Vdash \forall x\;\varphi(x) \iff \text{for all $i\le j\le n$ and $a\in \tau_j$ we have }\tau,j\Vdash \varphi(a)$
$\tau,i\Vdash \exists x\;\varphi(x) \iff \text{either $i=n$ or for some $a\in \tau_{i+1}$ we have }\tau,i+1\Vdash \varphi(a)$

Note that for $i\ge j$ we have $\tau, j\Vdash \varphi\Rightarrow \tau,i\Vdash \varphi$. And note that for a formula $\varphi$ of logical depth $l$ and $i\le \mathsf{len}(\tau)-l$ we couldn't have $\tau,i\Vdash \varphi$ and $\tau,i\Vdash \lnot \varphi$ at the same time.
Essential idea here is that a fulfillment sequence is a finite approximation for a first-order model.
Let us say that a fulfulment sequence $\tau$ supports set of formulas $\Gamma$ if the following holds:

For any  $\varphi(\vec{x})$ that is a subformula of formula from $\Gamma$, any $i\le \mathsf{len}(\tau)$, and $\vec{a}\in \tau_i$ we have $i\Vdash \varphi(\vec{a})\lor\lnot\varphi(\vec{a})$.
We have $t(\vec{a})\in \tau_{i+1}$, for any term $t(\vec{x})$ appearing in $\Gamma$, any $i< \mathsf{len}(\tau)$, and $\vec{a}\in \tau_i$.

The following lemma connects finite deductions with fulfilement sequences:
Lemma 1. Suppose $\Gamma(x)$ is a sequent of logical depth $l$, $P$ is a deduction of $\Gamma$ in Tait calculus (with cuts) of the depth $k$, and $\tau$ is a fulfilment sequence that supports formulas appearing in $P$. Then for any $i$ between $k$ and $\mathsf{len}(\tau)-l$ and $\vec{a}\in\tau_i$ we have $\tau,i\Vdash \bigvee\Gamma(\vec{a})$
Proof. By induction on structure of $P$.
Suppose we have some finite sets of formulas $\Gamma\subseteq \Delta$ that are closed under subformulas. And suppose that we have some long enough fulfilment sequence $\tau$ that supports $\Gamma$. The key construction that we need is the construction of a shorter $\tau'$ such that $\tau'$ supports $\Delta$, $\mathfrak{M}_\tau=\mathfrak{M}_{\tau'}$, and for some function $f\colon \{0,..,\mathsf{len}(\tau')\}\to \{0,..,\mathsf{len}(\tau)\}$ we have

$\tau'_i\subseteq \tau_{f(i)}$,
$\tau',i\Vdash \varphi(\vec{a})\iff \tau,f(i)\Vdash \varphi(\vec{a})$, for any $\varphi(\vec{x})\in \Gamma$ and $\vec{a}\in \tau'_i$.

Let $s$ be the sum of lengthes of formulas from $\Delta$ and $k$ be the least number such that for any $\varphi\in \Delta$ and it's depth $k$ subformula $\psi$ we have $\psi\in \Gamma$. It would be always possible to construct $\tau'$ of the length $n$ as long as $\tau$ had the length $\ge P(n,s)\uparrow \uparrow k+1$, for some polynomial $P$. I'll skip technical details of construction of $\tau'$ from $\tau$.
Now our goal would be to construct long enough fulfulment sequence $\tau$ that would fulfill and support the set of all axioms of $T=\mathsf{Graham}+\forall x (x<g_{64})$. Let us construct finite sets of naturals $A_0=\{0\}$ and $$A_{i+1}=A_i\cup \{\max(0,a-1)\mid a\in A_i\}\cup \{\min(t(\vec{a}),g_{64})\mid \vec{a}\in \tau_i\text{ and }t(\vec{x}) \text{ occurs in axioms of $T$}\}.$$ Let $n$ be the last step so that $A_{n}\ne [0,g_{64}]$ and let $s$ be the least so that $[s,g_{64}]\subseteq A_n$. The model $\mathfrak{M}_\tau$ is the model with the domain $[0,s]$ that is obtained from $\mathbb{N}$ by collapsing all numbers $>s$ to $s$. We put $\mathsf{len}(\tau)=n$ and $\tau_i=(A_i\cap [0,s))\cup\{s\}$. It is easy to see that

$\tau$ supports the set of axioms of $T$
$\tau,i\Vdash \varphi$, for any $i\le \mathsf{len}(\tau)$ and axiom $\varphi$ of $T$
$\mathsf{len}(\tau)>\log_2(\log_2(g_{64}))$.

Finally assume for a contradiction that $P$ is a proof of the sequent $\lnot \mathsf{Graham},\exists x \lnot(x<g_{64})$ of the length $k$, where $k\le(\log_2^*(g_{64}))^{1/N}$. Then from $\tau$ constructed above we construct $\tau'$ so that

$\tau'$ supports the set of all formulas occuring in $P$
$\tau',i\Vdash \varphi$, for any $i\le \mathsf{len}(\tau)$ and axiom $\varphi$ of $T$
$\mathsf{len}(\tau')\ge k+l$, where $l$ is the logical depth of $\bigvee \lnot \mathsf{Graham}\lor \exists x \lnot(x<g_{64})$.

We get to a contradiction since from Lemma 1 should have that $\tau',k\Vdash \bigvee \lnot \mathsf{Graham}\lor \exists x \lnot(x<g_{64})$ but at the same time from 2. we have $\tau',k\Vdash \bigwedge \mathsf{Graham}\land \forall x (x<g_{64})$. And the latter is impossible since $k$ is too far away from the end of the sequence $\tau'$.
[1] J.E. Quinsey, "Some problems in logic: Applications of Kripke's Notion of Fulfilment", PhD thesis, St. Catherine's College, Oxford, 1980, https://arxiv.org/abs/1904.10540