Is Hilbert's epsilon calculus a conservative extension of intuitionist logic?

Question

Epsilon calculus is an extension to First Order Logic, by including the epsilon operator. The epsilon operator works as a sort of choice operator, which selects a witness for each true existential statement. I'll notate the epsilon operator as $\mathcal{E}$, and this operator can be fully axiomatized by asserting the following, for each predicate $P$. $$\exists xP(x) \iff P(\mathcal{E}xP(x))$$

Although at first $\mathcal{E}$ seems like it would imply Choice over $\sf{ZF}$ set theory, this only works if you allow Specification instances which mention $\mathcal{E}$. Counter-intuitively, so long as you do not allow $\mathcal{E}$ to be mentioned in any axioms of your theory (other than the basic axioms of First-Order Logic), then its inclusion is a conservative extension to classical logic. The proof is informative for why this happens.

Let $\phi$ be any sentence unprovable over classical FOL, restricted so that $\phi$ does not mention the $\mathcal{E}$ operator. It follows that $\neg\phi$ is consistent, so there exists a countable model $M$ for which $M\vDash\neg\phi$. Fix some wellordering of $M$, and select some distinguished object $m_0\in M$, then define $\mathcal{E}xP(x) = \min\{x\in M : M \vDash P(x)\}$ using the value $\mathcal{E}xP(x)=m_0$ in case no such $x$ exists. It's clear that $\mathcal{E}$ obeys the axioms for epsilon calculus, and we still do not prove $\phi$, hence epsilon calculus is a conservative extension to classical FOL.

My question is in the title: is the epsilon calculus also a conservative extension for intuitionist logic? That is, is there any instance of the Law of Excluded Middle which the epsilon can be used to prove, which is not already constructively valid?

My initial suspicion was that we will still have a conservative extension. After all, it seems like the epsilon is just a glorified way to express existential instantiation. In fact, if we restrict the usage of the operator only to unary formulae, such that it only ever constructs constant expressions, then we really do get a conservative extension for that exact reason. However, you can also use the operator to define functions, so the situation is not so simple.

I tried to adapt the Diaconescu-Goodman-Myhill Theorem to use $\mathcal{E}$ in place of a choice function, but the adaptation seems to fail. The proof relies on assuming the rule $(\forall xA(x)\iff B(x)) \implies \mathcal{E}xA(x) = \mathcal{E}xB(x)$ for all propositions $A,B$, but this isn't a necessary feature of Epsilon calculus. That said, the failure of this rule means the situation is much more complex. I don't even know how to construct non-classical models, so I have no clue how to proceed.

Answer 1

It is not a conservative extension to intuitionist logic, and in fact the Law of Excluded Middle succeeds catastrophically: while working over intuitionistic Peano Arithmetic (Heyting Arithmetic, $\sf{HA}$), we get LEM for all arithmetical sentences. More generally, the set of formulae obeying Double Negation Elimination (DNE) will be closed under quantification.

The reason this happens, ultimately, is because we can use any nonconstructive existence statement to prove a nearly identical constructive existence statement. The implication $\neg\neg\exists x, P(x) \implies \exists x, \neg\neg P(x)$ is known to be unprovable constructively, but this implication is provable over epsilon calculus. Since we know $\exists x, P(x) \implies P(\mathcal{E}[x,P(x)])$, then via contraposition we get the following implication chain. $$\neg\neg \exists x, P(x) \implies \neg\neg P(\mathcal{E}[x,P(x)]) \implies \exists x, \neg\neg P(x)$$

This is already enough to show we don't have a conservative extension.

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To get our stronger result, we shall prove that every formula over Heyting Arithmetic is "regular", meaning it obeys Double Negation Elimination. Notice that if $P$ is regular, then the rule established previously simplifies to $\neg\neg \exists x, P(x) \implies \exists x, P(x)$, hence the regular formulae are closed under existential quantification. Even without epsilon calculus, the regular formulae are already closed under negation, implication, conjunction, and universal quantification. This follows from the intuitionistic theorems below. $$\begin{align} \neg\neg\neg P &\implies \neg P \\ \neg\neg(P\implies Q) &\implies (P\implies \neg\neg Q) \\ \neg\neg(P\land Q) &\implies (\neg\neg P \land \neg\neg Q) \\ \neg\neg\forall x, P(x) &\implies \forall x, \neg\neg P(x) \end{align}$$

Finally, we prove that the regular formulae are closed under disjunction. The strategy for proving this is hopefully obvious: define the proposition $R(n)$ to be $(n=0\land P)\lor (n=1\land Q)$, then we see $(P\lor Q) \iff \exists(n<2), R(n)$, and consequently: $$\begin{align} \neg\neg (P\lor Q) &\iff \neg\neg \exists(n<2), R(n) \\ &\implies \exists(n<2), \neg\neg R(n) \\ &\implies (\neg\neg R(0))\lor (\neg\neg R(1)) \\ &\implies (\neg\neg P)\lor(\neg\neg Q) \end{align}$$

On the second line above I implicitly assume $n<2 \iff \neg\neg(n<2)$, but this is a theorem of $\sf{HA}$ so it's fine. More generally, Heyting Arithmetic proves $\neg\neg(x=y)\implies x=y$ via intuitionistic induction, so equality is regular, and thus all atomic formulae are regular. It now follows that every formula is regular, via metatheoretic induction on the length of the formula. Consequently we have Double Negation Elimination for every formula, and in light of the intuitionistic theorem $\neg\neg(P\lor \neg P)$, we also have the Law of Excluded Middle.