doubly negated intuitionistic formulas

Question

What is an example of some doubly negated statement in intuitionistic logic that is not equivalent to its classical version?

Some background: for every statement $P$ we can find a classically equivalent statement $Q$ that is valid classically iff it's valid intuitionistically. When $P$ does not have quantifiers, you can take $Q$ to be $\lnot \lnot P$. In general it's a bit harder.

So there are some statements $P$ such that $\lnot \lnot P$ is classically valid but not intuitionistically valid. What's an explicit example?

I guess in these situations you can add $\lnot P$ as an axiom and not arrive at a contradiction intuitionistically, but there would be a classical contradiction.

For example, $$\lnot \lnot \forall x\ (x=0 \lor x\ne 0)$$ in the real numbers might work, because its negation is $$\lnot \forall x\ (x=0 \lor x \ne 0)$$ which is not contradictory intuitionistically, I think. How do you prove that? Or what's another example if this one does not work?

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Edited: to make my question clearer. I will use sequent calculus as in the current answer, so $LJ$ is the intuitionistic system and $LK$ is the classical one.

For any formula $P,$ let $P^{\lnot\lnot}$ be a formula that is classically equivalent to $P$ and such that $\newcommand\intui{\vdash^{LJ}} \newcommand\classi{\vdash^{LK}} \intui P^{\lnot \lnot}$ iff $\classi P^{\lnot \lnot}$.

For simple formulas, $P^{\lnot\lnot}$ is just $\lnot \lnot P.$ For more complex formulas the double negation outside is not enough. So it is possible to have a formula $P$ with $\classi P$ but not $\intui \lnot \lnot P$. Thus $\lnot P$ is consistent intuitionistically, but not classically.

I was looking for an example of such $P$. In order to do that, I wanted a not purely logical example, so I searched inside a theory $\Gamma$, thinking that $\Gamma \classi P$ iff $\Gamma \intui P^{\lnot \lnot}.$ I was corrected (see answer below), because $\Gamma$ must also be changed. Even then, one can find examples of theories where $\Gamma \classi P$ but not $\Gamma \intui \lnot \lnot P$ (one is my example with a theory of the real numbers that allows it, according to the answer; I will have to study that).

In that case $\Gamma, \lnot P$ is consistent intuitionistically but not classically, and I wanted to know how to interpret that, with some example.

Answer 1

You ask multiple different questions. I will go through them instance-by-instance. Since I wasn't sure what you mean by the word valid (as in when you say that some statement is "classically valid"): I interpreted it as provability in the logic in question under some implicitly specified theory. If you meant something else, let me know.

What is an example of some doubly negated statement in intuitionistic logic that is not equivalent to its classical version?

Clearly, every double-negated sentence is classically equivalent to its non-double-negated version. However, any non-provable instance of the Law of Excluded Middle yields an example of a double-negated sentence that the intuitionistic theory in question won't prove equivalent to its classical version.

For example, let $P(x)$ denote "$x$ codes a proof of a contradiction in Heyting arithmetic". Then, by Gödel's incompleteness theorem, if Heyting arithmetic is $\omega$-consistent, it fails to prove both $\exists x. P(x)$ and $\neg \exists x. P(x)$. Since Heyting arithmetic has the disjunction property, it won't prove $\exists x. P(x) \vee \neg \exists x.P(x)$ either.

However, Heyting arithmetic does prove $\neg\neg ((\exists x. P(x)) \vee \neg \exists x. P(x))$. Indeed, $\neg \neg (A \vee \neg A)$ constitutes an intuitionistic tautology for any sentence $A$. The proof goes as follows. Since $A \rightarrow (A \vee \neg A)$ and $\neg A \rightarrow (A \vee \neg A)$ both hold, the contrapositives $\neg (A \vee \neg A) \rightarrow \neg A$ and $\neg (A \vee \neg A) \rightarrow \neg\neg A$ hold as well. So $\neg (A \vee \neg A)$ implies a contradiction, and consequently $\neg \neg (A \vee \neg A)$ holds.

Some background: for every statement P we can find a classically equivalent statement Q that is valid classically iff it's valid intuitionistically.

The double-negation translation gives us a sentence $S^{\neg\neg}$ so that $S \leftrightarrow S^{\neg\neg}$ holds in classical logic, and $\Gamma \vdash_{LK} S$ in classical logic precisely if $\Gamma^{\neg\neg} \vdash_{LJ} S^{\neg\neg}$ in intuitionisitic logic.

Alas, intuitionistically the theory $\Gamma^{\neg\neg}$ may have very little to do with the theory $\Gamma$. For example, if $\Gamma$ contains axioms of the form $(\forall x. L(x)) \rightarrow Q$ for atomic predicates $L,Q$, these will get translated in $\Gamma^{\neg\neg}$ as $(\forall x. \neg\neg L(x)) \rightarrow \neg\neg Q$. And the former need not imply the latter intuitionistically, nor does the latter need to imply the former intuitionistically.

So there are some statements P such that ¬¬P is classically valid but not intuitionistically valid.

Again, I'm not quite sure what you mean by valid here.

If you can find $P$ such that $\Gamma$ entails $\neg\neg P$ classically, but not intuitionistically, then you can dispense with the double-negation altogether: $\Gamma$ entails $P$ classically (since $P \leftrightarrow \neg\neg P$ holds classically), and $\Gamma$ does not entail $P$ intuitionistically (since $P \rightarrow \neg\neg P$ holds intuitionistically).

Really, you should say: there may be some statements $P$ such that $P$ is classically valid ($\Gamma \vdash_{LK} P$?) but not intuitionistically valid ($\Gamma \not\vdash_{LJ} P$?).

I guess in these situations you can add ¬P as an axiom and not arrive at a contradiction intuitionistically, but there would be a classical contradiction.

You seem to be using an inference along the following lines: if $\Gamma \vdash_{LK} P$ but $\Gamma \not\vdash_{LJ} P$, then $\Gamma \cup \{ \neg P \}$ is consistent in intuitionistic logic. This kind of reasoning does not work. For example, the empty theory satisfies $\vdash_{LK} A \vee \neg A$ for any formula $A$, but of course the theory $\{\neg (A \vee \neg A)\}$ is inconsistent even in intuitionistic logic.

That said, while your reasoning to establish it was not valid, the phenomenon does happen. For example, the theory $\{ \neg \forall x. M(x) \vee \neg M(x) \}$ in the language consisting of one predicate symbol $M$ is consistent intuitionistically. One can show this either by constructing a model (such as a Heyting-valued model, or a Kripke model, or a topological model), or explain by a combinatorial argument why $\neg \forall x. M(x) \vee \neg M(x) \vdash \bot$ has no cut-free proof tree in the intuitionistic sequent calculus LJ. I'll do the latter.

We define a rut as a sequent of one of the following 5 forms:

1. $M(t_1),M(t_2),\dots,M(t_n),\neg \forall x. M(x) \vee \neg M(x),\neg \forall x. M(x) \vee \neg M(x), \dots, \neg \forall x. M(x) \vee \neg M(x) \vdash \bot$,
2. $M(t_1),M(t_2),\dots,M(t_n),\neg \forall x. M(x) \vee \neg M(x),\dots, \neg \forall x. M(x) \vee \neg M(x) \vdash M(s)$,
3. $M(t_1),M(t_2),\dots,M(t_n),\neg \forall x. M(x) \vee \neg M(x), \dots, \neg \forall x. M(x) \vee \neg M(x) \vdash \neg M(s)$
4. $M(t_1),M(t_2),\dots,M(t_n),\neg \forall x. M(x) \vee \neg M(x), \dots, \neg \forall x. M(x) \vee \neg M(x) \vdash M(s) \vee \neg M(s)$
5. $M(t_1),M(t_2),\dots,M(t_n),\neg \forall x. M(x) \vee \neg M(x), \dots, \neg \forall x. M(x) \vee \neg M(x) \vdash \forall x. M(x) \vee \neg M(x)$

such that $s \not\in \{t_1,t_2,\dots,t_n\}$.

Now, we have the following simple observations:

1. If you apply a rule of the intuitionistic sequent calculus LJ to a rut, then you get another rut. For example, if you have $M(t_1),\neg \forall x. M(x) \vee \neg M(x) \vdash \forall x. M(x) \vee \neg M(x)$, and you apply the $\forall R$ rule, then you end up with $M(t_1),\neg \forall x. M(x) \vee \neg M(x) \vdash M(s) \vee \neg M(s)$ for some $s \neq t_1$.

2. A rut cannot conclude any branch of a proof. To conclude a branch, you'd have to use an axiom rule, but that requires the same formula to occur on both the left and the right of the turnstile. This clearly never happens in a rut.

These together imply that a rut has no intuitionistic proof: to conclude the proof, we'd somehow have to get out of the rut (2), but we cannot do that using intuitionistic rules (1).

Since $\neg \forall x. M(x) \vee \neg M(x) \vdash \bot$ is a rut, it also has no intuitionistic proof, so the theory $\{ \neg \forall x. M(x) \vee \neg M(x) \}$ in the language consisting of one predicate symbol $M$ is consistent intuitionistically. It is of course not consistent classically (and as you'd expect, contraction on the right $CR$, a rule that the intuitionistic calculus LJ does not contain, but the classical calculus LK does contain, immediately yields a non-rut from a rut).

How do you prove that [$\neg \forall x. x = 0 \vee \neg x = 0$ is not contradictory for intuitionistic real numbers]?

I don't know whether this can happen for intuitionistic (say Dedekind) reals. But there are contexts where it happens: $\neg \forall x. x = 0 \vee \neg x = 0$ is satisfied e.g. by the real line of Smooth Infinitesimal Analysis. A direct combinatorial proof showing the consistency of Smooth Infinitesimal Analysis, resembling the one above, would be prohibitively difficult to write down. Instead, you usually have to construct a model where the interpretation of $\neg \forall x. x = 0 \vee \neg x = 0$ actually holds. The first chapter of Moerdijk-Reyes will give you an idea of how it's done.