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$\forall x \neg \neg \phi(x) \to \neg \neg \forall x \phi(x)$ is not intuititionistically valid. I can come up with a complicated Kripke counter-model as follows:

Let there be a countably infinite number of related states $w_0 \le w_1 \le w_2…$ and objects $a_0, a_1, a_2, …$ such that $\phi(a_0), \phi (a_1),… \phi(a_n)$ hold at each state $w_n$, and $\phi(a_{n+1}), \phi(a_{n+2}),…$ do not hold at $w_n$. So, since for every object $a_i$ at every accessible state from $w_0$ there is an accessible state where $\phi(a_i)$ holds, we have $\forall x \neg \neg \phi(x)$. However, there is no accessible state from $w_0$ where $\phi(x)$ holds for all $a_i$. Thus, we do not have $\neg \neg \forall x \phi(x)$.

This seems to be a really inefficient counter-model, but I can’t seem to figure out a simpler one in Kripke semantics. Is there a simpler approach?

PW_246
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  • a better way of writing "we do not have not not for all x phi x" would be "there exists x for which not phi x" – Snared Jul 05 '23 at 23:57
  • That’s the whole point; in fact, $\forall x \neg \neg \phi$ is still intuitionistically equivalent to $\neg \exists x \neg \phi$. – PW_246 Jul 06 '23 at 00:04
  • Sidenote to quickly buy into it: For $\phi(x):=\psi(x)\lor\neg\psi(x)$, we have $\forall x.\neg \neg \phi(x)$ as a logical theorem. At the same time, for $\psi(x)$ representing the halting problem, $\neg \forall x.\phi(x)$ is consistent with $\mathsf{HA}$, and so cannot be negated by it. – Nikolaj-K Aug 01 '23 at 23:51
  • @Nikolaj-K thanks for the reply. Is $\neg \forall n \phi(e)$ supposed to be $\neg \forall e \phi(e)$? Also, I’m not seeing how this pertains directly to the question, but it’s insightful either way. – PW_246 Aug 01 '23 at 23:51
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    Yes, there's just one variable, I first had your $x$ but the other names are probably more common for the indices for this computation predicate. I edited it back to $x$ now. I write Sidenote since it doesn't answer the question in the way you asked it. But it gives the imho quickest way to justify why the double-negation shift isn't permissible. Thought that's what you're really after. – Nikolaj-K Aug 01 '23 at 23:54
  • @Nikolaj-K that makes sense. I would say I understand why the double-negation shift is impermissible from a constructivist perspective. I just am not sure how to give a better counter-model in Kripke semantics. – PW_246 Aug 02 '23 at 04:05

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There isn’t really a nicer model. Any model with finitely many states will fail. To see this, note that maximal elements are “dense” in the finite state case (every element is less than a maximal element), and note that a maximal element necessarily models $\forall x \phi(x)$, since maximal states model classical logic.

A model which has infinitely many states but only finitely many objects will also fail, though the reasoning here is somewhat more complicated.

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