If x does not occur free in A then (∃x)A→A an absolute theorem schema?

Question

My Thoughts:
Logically if (∃x)A is false then (∃x)A→A is true. But if (∃x)A is true. Then A
must also be true. So it seems this is an absolute theorem schema.
But how can I give a Hilbert or equational style proof of this?

Answer 1

A reasonably short proof of this is that whenever $x$ does not occur free in $Q$, $(\exists x P) \to Q$ is logically equivalent to $\forall x (P \to Q)$.

So in particular, $(\exists x A) \to A$ is equivalent to $\forall x (A \to A)$.

Now $A \to A$ is a tautology, so universally quantifying over it results in a tautology.

Answer 2

There's no contradiction here and I offer an intuitive answer. You misused material conditional connective when you're seemingly logically and thus confidently saying "...then (∃x)A→A is true. But if (∃x)A is true." The previous if/then is a conditional truth functional connective which is true (we say the if/then argument is valid in strict English), and the reason why it's true is simply the antecedent is false per material condition definition.