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I am trying to prove the statement in the form:

$\exists x \in S, \forall y \in \emptyset, P(x)$

where $P(x): x > y$ and $S$ is an arbitrary non-empty set.

My proof looks something like:

Let $x =$ some number in set S. Let $y \in \emptyset$. My intuition is that since we cannot check whether $x > y$ since $y$ is an empty set, the statement is vacuously true. My questions is, how do I write this "intuition" in a rigorous way?

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    $\begingroup$ What you have written is already rigorous. $\endgroup$ – Kavi Rama Murthy 2 days ago
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You could restate the claim as $\exists x\in S\neg\exists y\in\emptyset\neg P(x)$, which is trivial for each $x\in S$.

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This reduces to: $\exists x \colon y \in \{\} \implies P(x)$.

But $y \notin \{\}$.

The truth table of material implication/conditionals ('IF') definitionally maps False to True.

QED.

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  • $\begingroup$ I think you mean $\exists x,\,\forall y,\,(y\in\{\}\Rightarrow P(x))$ . $\endgroup$ – Gae. S. 2 days ago

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