Proof of the negation of a universal statement

Proof

We show that the two propositions have the same truth value for each set $X$ and each predicate $P$ on $X$.

1. Let $X$ be a set and let $P(x)$ be a predicate on $X$. For each $a \in X$, the proposition $P(a)$ is true or false.
2. By the definition of negation, $\neg \bigl( \forall x \in X, \, P(x) \bigr)$ is true exactly when $\forall x \in X, \, P(x)$ is false.
3. By the definition of the universal quantifier, $\forall x \in X, \, P(x)$ is false exactly when it is not the case that $P(a)$ is true for each $a \in X$.
4. This means that there is some $a \in X$ for which $P(a)$ is not true. By step 1, such a $P(a)$ is false.
5. By the definition of negation, $P(a)$ is false exactly when $\neg P(a)$ is true.
6. By steps 2 to 5, $\neg \bigl( \forall x \in X, \, P(x) \bigr)$ is true exactly when $\neg P(a)$ is true for at least one $a \in X$.
7. By the definition of the existential quantifier, this is exactly when $\exists x \in X, \, \neg P(x)$ is true.
8. So the two propositions are true in the same cases, and false in the same cases. Hence $\neg \bigl( \forall x \in X, \, P(x) \bigr) \equiv \exists x \in X, \, \neg P(x)$.