Show the following derivability fact: $\exists x : (P(x) \land x = c),\quad \forall x(P(x) \implies Q(x)) \vdash Q(c)$

Question

I've come across the following question:

Show the following two derivability fact:

$\exists x : (P(x) \land x = c),\quad \forall x(P(x) \implies Q(x)) \vdash Q(c)$

Now I'm not really sure where to start here. The solution provided uses a tableau, but I strongly dislike those, so am looking for other ways to show the same thing.

Answer 1

With Natural Deduction :

1) $∃x(P(x) ∧ x=c)$ --- premise

2) $∀x(P(x) \to Q(x))$ --- premise

3) $P(a) ∧ a=c$ --- from 1) for $\exists$-elim (here we start the subproof)

4) $P(a)$ --- from 3) by $\land$-elim

5) $a=c$ --- from 3) by 4\land$-elim

6) $P(a) \to Q(a)$ --- from 2) by $\forall$-elim

7) $Q(a)$ from 4) and 6) by $\to$-elim [aka : Modus Ponens]

8) $Q(c)$ --- from 5) and 7) by $=$-elim [aka : substitution of equality for formulas : $x = y → (φ[z/x] → φ[z/y]$)](here ends the sub-proof)

9) $Q(c)$ --- from 1) and 3)-6) by $\exists$-elim.