Very Basic Logic Question

Question

Given a set $S=\{-1,0,-5,-4\}$.Then is the following proposition true? $\forall x, (x>0 \implies x^2>0)$.

Answer 1

Since no element of $S$ is greater than zero, the implication is vacuously true.

$$\forall x\in S(x>0\to x^2>0) \iff \neg\exists x\in S(x>0\wedge x^2\not\gt 0)$$

There does not exist a counter example; of an element in $S$ which is both greater than zero but having a square not greater than zero.

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Note that this is also (vacuously) true: $\forall x\in S(x>0\to x^2< -1)$