A question about $(\exists x: p(x) \supset q(x)) \equiv ((\forall x: p(x)) \supset (\exists x: q(x)))$

Question

Conider the usual logical connectives $\wedge$ for "and", $\vee$ for "or", $\supset$ for material implication, $\equiv$ for material equivalence, $\neg$ for "not" et cetera.

We all know and use laws like $(p \supset q) \equiv (\neg q \supset \neg p)$ and $(\exists x: \neg p(x)) \equiv \neg(\forall x: p(x))$ et cetera.

In many introductions to logic there are many more laws to be found. Consider the following: $(\exists x: p(x) \supset q(x)) \equiv ((\forall x: p(x)) \supset (\exists x: q(x)))$

Has anyone ever used (or seen used) this particular law in any interesting mathematical context (outside of logic) and can provide a concrete example where it helped in an inference?

Answer 1

The inference "licensed" by :

$∃x(p(x)⊃q(x))≡(∀xp(x)⊃∃xq(x))$

is quite "natural" indeed ...

Consider the following interpretation for the predicates : $p$ stands for "... is a Philosopher" and $q$ for "... is an enQuirer".

From left to right :

if we know that (LHS) :

there is at least one person such that, if he is a Philosopher, then he is an enQuirer,

we are licensed to assert that (RHS) :

if all the people in the universe are Philosopher, then there is at least one person which is an enQuirer.