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Suppose $Pxy$ is a $2$-ary predicate and $\phi$ is just some fixed arbitrary predicate form.

  1. $\forall x [ (\exists y Pxy) \longrightarrow \phi ]$

  2. $\forall x \exists y [Pxy \longrightarrow \phi]$

Is the only difference between 1. and 2. in the case where $\exists y$ is not true? i.e. We may be in an interpretation $\mathcal{I}$ where the universe/domain of discourse may be empty, so does that then make 1. true and 2. false?

So it seems to me that if we restrict ourselves to interpretations $\mathcal{I}$ whose domain of discourse is not empty then 1. and 2. are logically equivalent$^1$.


Remarks: $^1$I'm thinking of the interpretations in the context of first-order logic, but what about second-order logic or other higher orders of logic? (Though I have not studied anything beyond first order logic, but maybe someone can say something about this? But this probably is another question in its own right)

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Hint

The two are not equivalent.

Consider an interpretation with domain $\mathbb N$ and let $\phi := (0=1)$ (or any other formulas which is False).

Then interpret $Pxy$ with $x < y$.

In this interpretation, 1. will be $∀x[∃y(x < y) ⟶ (0=1)]$, which is False, while 2. is $∀x∃y[(x < y) ⟶ (0=1)]$, which is True.

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  • $\begingroup$ Thanks for this! I just realised after reading your hint that what I wrote above is vacuously true for both 1. and 2 and because of that, after thinking about these predicate forms I happen to have another question in relation to them. See math.stackexchange.com/questions/4143026/… if interested. $\endgroup$
    – tcmtan
    May 18 at 11:12

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