Existential Instantiation

Context
Predicate Logic.

Theorem
$$\exists x: P \left({x}\right), P \left({\mathbf{a}}\right) \Longrightarrow y \vdash y$$

If, from our universe of discourse, an arbitrarily selected object $$\mathbf{a}$$ which has the property $$P$$ implies a conclusion $$y$$, and it is known that there does actually exists an object that has $$P$$, then we may infer $$y$$.

Some authors call this the Rule of Existential Elimination.

When using the rule of existential instantiation:

$$\exists x: P \left({x}\right), P \left({\mathbf{a}}\right) \Longrightarrow y \vdash y$$

the instance of $$P \left({\mathbf{a}}\right)$$ is referred to as the typical disjunct.

Proof
This is an extension of the Rule of Or-Elimination.

The propositional expansion of $$\exists x: P \left({x}\right)$$ is:

$$P \left({\mathbf{X}_1}\right) \lor P \left({\mathbf{X}_2}\right) \lor P \left({\mathbf{X}_3}\right) \lor \ldots$$

We know that any arbitrarily selected $$\mathbf{a}$$ with the property $$P$$ implies $$y$$.

From this we can infer that all such $$\mathbf{a}$$ which have that property imply $$y$$.

This is equivalent to the step in the Rule of Or-Elimination in which we need to prove that both disjuncts lead to the same conclusion.

The fact that we only need one of them in fact to be true is quite enough to draw the conclusion that $$y$$ is true.

In this context, we are assured by the statement $$\exists x: P \left({x}\right)$$ that at least one such disjunct in the above propositional expansion is true.

Thus the conclusion follows, and the result is proved.