I'm trying to solve this simple natural deduction problem:
$\{∀x~(p(x) \to q(a)), ∀y~\lnot q(y)\} \vdash_{\small ND} \lnot p(a)$
I started out by stating the premisses and the assumption, which is $p(a)$. I used $p(a)$ in the first premise with the (→ E) rule to get $q(a)$. Here's the problem, i got a $q(a)$ and a $∀y~¬q(y)$ which turns into $¬q(b)$? since i already turned x to a.
The question is, can i turn the y to a aswell? If so it should be pretty easy to prove this. Or am i just completely off track?