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Of course, in classical logic, if you have $\forall x \forall y (P(x) \lor Q(y))$, then you can conclude $(\forall x, P(x)) \lor (\forall y, Q(y))$: if you didn't have that, then there would be $x$ such that $\lnot P(x)$ and $y$ such that $\lnot Q(y)$, and then that contradicts the assumption.

On the other hand, in intuitionistic logic, I don't think that necessarily holds. My counterexample is to work in the topos of sheaves on $\mathbb{R}$, let the domain of discourse be $\mathbb{N}^+$, and define $P(n) := (-\infty, \frac{1}{n})$ and $Q(n) := (-\frac{1}{n}, \infty)$. Then for any $m, n \in \mathbb{N}^+$, $P(m) \cup Q(n) = \mathbb{R}$, so the hypothesis should also be valid for any locally constant functions $m, n : U \to \mathbb{N}^+$, i.e. sections of the constant sheaf $\mathbb{N}^+$. On the other hand, $\bigcap_m P(m) = (-\infty, 0)$ and $\bigcap_n Q(n) = (0, \infty)$, and the union of those two is $\mathbb{R} \setminus \{ 0 \}$.

So, my question is to verify that this is a correct solution and I didn't miss anything silly. I would also be interested to see whether a counterexample could be constructed in terms of a Kripke frame, where no example immediately comes to mind for me.

Daniel Schepler
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  • Oh, by the way, I came up with this question while looking at: https://math.stackexchange.com/a/4640514/337888 – Daniel Schepler Feb 16 '23 at 17:59
  • Just in case you need reassurance: $(\forall x. \forall y. Px \vee Qy) \rightarrow (\forall x. Px) \vee (\forall y. Qy)$ is indeed not an intuitionistic tautology, and this is one of those nice cases where the syntactic/combinatorial proof is not difficult either, by considering possible derivations of $(\forall x. \forall y. Px \vee Qy) \vdash (\forall x. Px) \vee (\forall y. Qy)$ in the cut-free fragment of LJ. left. This is a good exercise for those who wish to hone their proof theory skills. – Z. A. K. Feb 16 '23 at 20:04
  • Hmm, so there's a version of cut-free sequent calculus which is complete for intuitionistic FOL with two unary predicates? Interesting. I remember seeing some mention of the possibility though when I self-studied sequent calculus I was concentrating on propositional logic. I also thought there was no hope of such a thing existing for e.g. FOL languages complex enough to express the group word problem, or NBG set theory. – Daniel Schepler Feb 16 '23 at 20:12
  • I'm saying one can analyze cut-free proofs in LJ and prove directly, using a combinatorial argument, that $(\forall x. \forall y. Px \vee Qy) \vdash (\forall x. Px) \vee (\forall y. Qy)$ cannot have an LJ-derivation (and is therefore not a theorem of intuitionistic logic). This does not imply the existence of a complete proof-search procedure that works for some particular fragment of monadic IFOL (indeed, there is no such thing: in the intuitionistic case just one unary predicate is enough to get you undecidability). – Z. A. K. Feb 16 '23 at 20:29
  • I guess I don't know enough about LJ in the first-order logic case to really know what you're trying to get at. Is it that this case of LJ no longer satisfies the subformula property so there is no longer a finite set of possible contexts to consider in a proof search, but there is still enough structure to it to exclude the possibility of this particular statement having an LJ-derivation? Or is it something else entirely? – Daniel Schepler Feb 16 '23 at 20:58

1 Answers1

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Your logic appears sound.

Another approach is as follows. We work in constructive set theory. Fix a one-element set $1$ and a proposition $P$. Define $P(x) = P$ and $Q(x) = \neg P$.

Consider $S = \{x \in 1 \mid P \lor \neg P\}$. Then $\forall x, y \in S, P(x) \lor Q(y)$. However, if we have $\forall x \in S (P(x))$, then we have $\neg \neg P$. And if we have $\forall y \in S (Q(y))$, then we have $\neg P$. So to assert $(\forall x \in S P(x)) \lor (\forall y \in S Q(y))$ is precisely to assert $\neg P \lor \neg \neg P$. Since $\neg P \lor \neg \neg P$ is not a constructive tautology, we have demonstrated that OP’s implication is not a constructive tautology.

If you want a specific Kripke frame where this fails, you can start with a Kripke frame where $\neg \neg P \lor \neg P$ fails for some $P$, and then trace backwards along my proof. One example involves a 3-element frame with elements $a, b, c$, where the relation is $x \leq y$ precisely when $x = c$. We then define $P$ to hold only at $b$. Then $\neg P$ cannot hold at $c$, since $P$ holds at $b \geq c$. Similarly, $\neg \neg P$ cannot hold at $c$, since $\neg P$ holds at $a \geq c$. Thus, $\neg \neg P \lor \neg P$ doesn’t hold at $c$.

We take $S(a) = 1 = S(b)$ and $S(c) = \emptyset$. We define $P(x)$ to hold at $b$ and fail at $a$, and we define $Q(x)$ to hold at $a$ and fail at $b$. Then we indeed have $\forall x \forall y (P(x) \lor Q(y))$ in our Kripke model, since once we have an $x$ and a $y$, we must be at stage $a$ or $b$. But at stage $c$, we don’t have either $\forall x P(x)$ or $\forall y Q(y)$.

Note that we defined $S, P(x), Q(x)$ in the preceding paragraph by following my constructive set theory construction in the presheaf topos corresponding to the Kripke frame. Kripke models are just models in the internal logic of a topos of presheaves on a poset.

Mark Saving
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