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I am researching Mint's paper: Intuitionistic Existential Instantiation and Epsilon Symbol (this is as far as I know unfinished work)

In intuitionistic logic, it is not difficult to prove that $$\exists x[P(x)] \to \exists y[\exists x[P(x)]\to P(y)]$$

However, I got interested in Hilbert's $\varepsilon$, a choice operator of some sort. This $\varepsilon$ creates a term of the form $\varepsilon x.P(x)$ where $P(x)$ is a formula with at least one free variable (we keep it at exactly one for simplicity). The term $\varepsilon x.P(x)$ is intuitively interpreted as "a term for which $P$ holds". It is formally accompanied by the so called "critical axiom": $$\exists x[P(x)]\to P(\varepsilon x.P(x))$$

In Mint's paper, he tries to establish a completeness result with respect to certain Kripke models. For this, he defines a different natural deduction system that can accompany the $\varepsilon$-terms. The rules of this system are the same for all axioms and propositional inference rules as well as for the $\forall$-introduction inference rule. The other three, $\exists$-intro, $\exists$-elim and $\forall$-elim are changed to the following:

$$\dfrac{\exists x[P(x)]}{P(\varepsilon x.P(x))}\exists\text{-elim}, \dfrac{t\downarrow \ P(t)}{\exists x[P(x)]}\exists\text{-intro}, \dfrac{t\downarrow \ \forall x[P(x)]}{P(t)}\forall\text{-elim}$$

Here $t\downarrow := \top$ if $t$ is a variable or a constant and if $t$ is an $\varepsilon$ term then $$\varepsilon x.P(x)\downarrow := \exists y[\exists x[P(x)]\to P(y)]$$

Now I am stuck. Mint assures that we still have $\exists x[P(x)] \to \exists y[\exists x[P(x)]\to P(y)]$. This is also what I believe to be true. Yet, after several hours of trying, I can't find the proof. Working bottum-up we first use a $\to$-intro and then we can only use the new $\exists$-intro (I think). But since we do not have a $t$ we need to use an $\varepsilon$-term. Doing this requires us to proof $\exists x[P(x)] \to \exists y[\exists x[P(x)]\to P(y)]$ itself yet again.

I therefore ask if someone can tell me if there is something wrong with my reasoning, with the system itself or something else.

I thank you in advance

Z. A. K.
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Tungsten
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  • Do they make that assurance? They say that if "there exists x s.t. P(x) implies the x s.t. P(x) is definable" in section 3, lemma 2, but this is by induction on derivations, and may no longer hold internally. – Soundwave Feb 15 '24 at 06:18
  • @Soundwave In that light, It may indeed be questioned. However, Shoji Maehara also makes this claim in A General Theory of Completeness Proofs on page 65 just below the start of the proof of property 6. – Tungsten Feb 15 '24 at 09:48
  • @Soundwave: I don.t think that's it. I don't see how one would carry out the derivation of Section 5 at all without ezplicitly deriving $\exists x[P(x)] \Rightarrow \exists y[\exists x[P(x)]\to P(y)]$ at some point. – Z. A. K. Feb 16 '24 at 23:47
  • @Z.A.K. and yet, I see no issue with their argument, and it does not seem to get better for the sequent calculus either. $\exists x[P(x)]\Rightarrow\varepsilon x[P(x)]\downarrow$ seems conservative but not derivable. – Soundwave Feb 16 '24 at 23:59
  • @Soundwave: There's no way for $\exists x[P(x)]\Rightarrow\varepsilon x[P(x)]\downarrow$ to be "conservative but not derivable": in Section 5, this is not a schema in $P$, but a single formula containing a fixed predicate $P$. This formula is either derivable in the system or not; in the latter case, adding it is certainly not conservative. – Z. A. K. Feb 18 '24 at 01:59
  • I have more thoroughly inspected Maehara's argument and can only conclude it's circular. The derivation of the sequent $\exists x[P(x)]\Rightarrow\epsilon x.P(x)\downarrow$ in the completeness proof presupposes it during property 6. – Soundwave Feb 18 '24 at 10:34
  • There is one key difference between Maehara and Mint that I noticed recently. Mint changes the $\exists \Rightarrow$ to $\exists_{\varepsilon} \Rightarrow$. Maehara on the other end, does not change $\exists \Rightarrow$ but instead adds the critical axiom to his system. Now in Maehara's system, it is possible to prove $\exists x[P(x)]\Rightarrow \varepsilon x[P(x)]\downarrow$ because we keep the "normal" $\exists \Rightarrow$. – Tungsten Feb 19 '24 at 13:52
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    @Tungsten: You should migrate this question to MathOverflow now. There are a few proof theorists there who don't contribute here, but might be willing to answer there. It's certainly a research-level question: it concerns derivability in a recentish, "experimental" system, and having an answer is crucial for correctness of said work. And you can mention that it got many upvotes and no answer here, despite the considerable bounty. – Z. A. K. Feb 25 '24 at 00:53

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