# quantifier order in monadic first-order logic

In (polyadic) first-order logic, for any sentence $\psi(x,y)$ with variables $x$ and $y$ free, we have $(\exists y)(\forall x)\psi(x,y)\vDash (\forall x)(\exists y)\psi(x,y)$ but not the reverse entailment. Is the reverse entailment true in monadic predicate logic, however? And if so, how would a proof go?

It strikes me as a little bizarre that I can't find a reference on this point, but perhaps the answer is obvious and I'm just not seeing it. (Heuristically, though, it would seem to be worth making explicit.)

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In monadic predicate logic, there are no atomic sentences with two free variables. What's an example of the entailment you're asking about? –  MikeC Nov 13 '11 at 4:22
Yes, that's why I didn't specify the sentences were atomic, though I guess I could have explicitly said molecular. I'm thinking of something simple (in this case it's easy to see the reverse entailment holds): take \psi(x,y) to be (Fx\to Gy). –  symplectomorphic Nov 13 '11 at 4:36

With a monadic predicate $P$, the sentence $(\forall x)(\exists y)(P(x)\iff P(y))$ does not entail $(\exists y)(\forall x)(P(x)\iff P(y))$. In fact, the former is logically valid but the latter fals in any structure where the interpretation of $P$ is neither the empty set nor the whole universe.