$\exists (p(x)\rightarrow q(a)) \vdash \forall y p(y) \rightarrow q(a)$

Question

$\exists x(p(x)\rightarrow q(a)) \vdash \forall y p(y) \rightarrow q(a)$

How can we prove this with natural deduction? I couldn't any similar example from books or web. Do you have any idea?

Answer 1

You need the Natural Deduction rules for quantifiers.

1) $∃x(p(x)→q(a))$ --- premise

2) $∀yp(y)$ --- assumed [a]

3) $p(x)$ --- from 2) by $∀$-elim

4) $p(x)→q(a)$ --- assumed [b] for $∃$-elim

5) $q(a)$ --- from 3) and 4) by $\to$-elim

6) $q(a)$ --- from 5) and 4) and 1) by $∃$-elim, discharging [b]

7) $∀yp(y) \to q(a)$ --- from 2) and 6) by $\to$-intro, discharging [a].