Natural deduction: negation of quantifiers

Question

How can I show that $\lnot \exists x P(x) \vdash \forall x\lnot P(x)$ ?

Because I want to show: $\lnot \exists x (P(x) \lor R(x)) \vdash \forall x \lnot R(x)$

My idea: maybe a proof by contradiction would work...

Answer 1

For $¬∃xP(x)⊢∀x¬P(x)$, it is enough to assume $P(x)$ :

1) $¬∃xP(x)$ --- premise

2) $P(x)$ --- assumed [a]

3) $∃xP(x)$ --- from 2) by $∃$-introduction

4) $\bot$ --- from 1) and 3)

5) $¬P(x)$ --- from 2) and 4) by $¬$-introduction, discharging [a]

6) $∀x¬P(x)$ --- from 5), by $∀$-introduction.