How to prove ~($\forall$x Q(x)) is logically equivalent to $\exists$x(~Q(x)) using natural deduction for first order logic

Question

I am thinking of assuming Q(x1) and then deriving to reach to a contradiction but I have not been able to do so.

Answer 1

Hint

1st part, using Natural Deduction :

1) $\exists x \lnot Qx$ --- premise

2) $\forall x Qx$ --- assumed [a]

3) $\lnot Qa$ --- assumed [b] from 1) by $\exists$-elim

4) $Qa$ --- from 2) by $\forall$-elim

5) $\bot$ --- contradiction : from 3) and 4)

6) $\bot$ --- from 1) and 3) and 5) by $\exists$-elim, discharging [b]

7) $\lnot \forall x Qx$ --- from 2) and 6) by $\lnot$-intro, discharging [a].

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2nd part : quite similar, using Double Negation.

1) $\lnot \forall x Qx$ --- premise

2) $\lnot \exists x \lnot Qx$ --- assumed [a]

3) $\lnot Qx$ --- assumed [b]

and so on ...