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)



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.

  • $\begingroup$ 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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