In How To Prove It?, Velleman explains that, in order to prove a statement with a goal of the form $\exists x P(x)$, we must declare an arbitrary variable $y$ and assign a value $a$ to it and use the assignment to prove the substitution instance $P(y/x)$ of the goal. Hence, assuming $\Gamma$ is a well-founded formula, we have:
- $\Gamma \rightarrow \exists x P(x)$
- $(\Gamma \land (y=a)) \rightarrow P(y/x)$
My question concerns the formal structure of such proof technique and the relation between statements 1 and 2.
My first attempt was this one: since $P(a)$ is logically equivalent to $\forall y[(y=a) \rightarrow P(y)]$, then $\Gamma \rightarrow P(a)$ is equivalent to $\Gamma \rightarrow \forall y[(y=a) \rightarrow P(y)]$. If $y$ does not occur in $\Gamma$, then the last statement is equivalent to $\forall y[\Gamma \rightarrow ((y=a) \rightarrow P(y))]$, which means $\forall y[(\Gamma \land (y=a)) \rightarrow P(y)]$.
Now, since $\Gamma \rightarrow P(a)$ implies $\Gamma \rightarrow \exists x P(x)$, it is fair to believe that the relation between statements 1 and 2 is $\forall y[(\Gamma \land (y=a)) \rightarrow P(y)] \rightarrow [\Gamma \rightarrow \exists x P(x)]$.
However, some members on this site informed me that this implication is a weaker version of the more general assertion $\forall y[[(\Gamma \land (y=a)) \rightarrow P(y)] \rightarrow [\Gamma \rightarrow \exists x P(x)]]$. In this case, $y$ appears as a free variable, whose quantification ranges upon the entire proof. However, I can't tell how it is possible to derive such stronger implication, especially because the formula $(\Gamma \land (y=a)) \rightarrow P(y)$ seems meaningless to me.