I have the formula's $\forall x \space P(x) \rightarrow Q(x)$ and $\forall x \space P(x) \rightarrow Q(y)$.
Solving one:
- $\forall x \space P(x) \rightarrow Q(y)$
- $\neg \forall x \space P(x) \vee Q(y)$ implication-elimination
- $\exists x \space \neg P(x) \vee Q(y)$ duality (don't know English word for it)
- $\exists x \space \neg P(x) \vee \exists x \space Q(y)$ free Existential quantification
- $\exists x \space (\neg P(x) \vee \space Q(y))$ distribution of $\exists x$ over $\vee$
- $\exists x \space (P(x) \rightarrow Q(y))$ implication-elimination
Could someone tell me if this is correct?
And as well, I keep thinking this should also work for the other formula, $\forall x \space P(x) \rightarrow Q(x)$, but then I have to make sure the second $x$ is not free. Can I just rename it?
Yes, in both cases, the variables $x$, $y$, respectively, are "free" (not bound, as the scope of the universal quantifier is limited to the predicate $P(x)$).
So each statement says precisely the same thing.
But yes, in your proof of the first statement, you'd need to rename the variable prior to step $4$, i.e., replace $x$ in $Q(x)$ by, say, $y$, or $z$ (i.e., by any variable other than $x$) so it remains free in subsequent steps of the proof.
I'd suggest that you approach the entire problem by first stating $$\forall xP(x) \rightarrow Q(x) \equiv \forall xP(x) \rightarrow Q(y)$$
And then follow with your proof, as is, since it will then cover both statements, since they are logically equivalent.