Universal Negative implies Particular Negative iff First Predicate is not Vacuous

Theorem
Consider the categorical statements:
 * $\mathbf E: \quad$ The universal negative: $\forall x: S \left({x}\right) \implies \neg P \left({x}\right)$
 * $\mathbf O: \quad$ The particular negative: $\exists x: S \left({x}\right) \land \neg P \left({x}\right)$

Then:
 * $\mathbf E \implies \mathbf O$

iff:
 * $\exists x: S \left({x}\right)$

Sufficient Condition
Let $\exists x: S \left({x}\right)$.

Let $\mathbf E$ be true.

As $\mathbf E$ is true, then by Modus Ponendo Ponens:
 * $\neg P \left({x}\right)$

From the Rule of Conjunction:
 * $S \left({x}\right) \land \neg P \left({x}\right)$

Thus $\mathbf O: \quad$ holds.

So by the Rule of Implication:
 * $\mathbf E \implies \mathbf O$

Necessary Condition
Let $\mathbf E \implies \mathbf O$.

Suppose:
 * $\neg \exists x: S \left({x}\right)$

that is, $S \left({x}\right)$ is vacuous.

From De Morgan's Laws: Denial of Existence:
 * $\forall x: \neg S \left({x}\right) \dashv \vdash \neg \exists x: S \left({x}\right)$

it follows that $\forall x: S \left({x}\right)$ is false.

From False Statement implies Every Statement:
 * $\forall x: S \left({x}\right) \implies \neg P \left({x}\right)$

is true.

So $\mathbf E$ holds.

Again, $\neg \exists x: S \left({x}\right)$.

Then by the Rule of Conjunction:


 * $\neg \left({\exists x: S \left({x}\right) \land \neg P \left({x}\right) }\right)$

That is, $\mathbf O$ does not hold.

So $\mathbf E$ is true and $\mathbf O$ is false.

This contradicts $\mathbf E \implies \mathbf O$ by definition of implication.

Thus $\exists x: S \left({x}\right)$ must hold.