# Natural deduction predicate logic

I'm trying to solve this simple natural deduction problem:

{∀x(p(x) → q(a)), ∀y¬q(y)} |-nd ¬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?

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Yes, you can infer $\neg q(a)$ from $\forall{y}:\neg q(y)$. – mjqxxxx Oct 22 '12 at 23:01

You can instantiate a universal quantifier with any constant (whether you've used it before in the deduction or not). For '$\forall$' means all, doesn't it?! So if you are given '$\forall y\neg Q(y)$', then anything you take will satisfy the condition expressed by '$Q$' -- in particular, the thing dubbed '$a$'.
As a footnote, it is probably not a great idea to think in terms of 'turning $x$ into $a$' or the like: what exactly could that mean? You are, to use the standard jargon, instantiating the quantifier -- i.e. moving from a general claim (and it is the whole quantifier/variable apparatus $\forall x ... x ...$ which expresses the generality) to an instance of it. And you do that by-removing-the-initial-quantifier-and-substituting-a-name-for-the-variable(s)-that-it-binds, which is best thought of as a single operation.