Certainly if P is false, (P&Q) cannot be true.

But how to prove this using natural deduction?

I'd propose as a direct proof the following derivation :

(1) ~P ( Premise )

(2) ~P v ~Q ( v - intro)

(3) ~ ( P & Q) ( DeMorgan)


1) $\lnot P$ --- premise

2) $(P \land Q)$ --- assumed [a]

3) $P$ --- from 2) by $(\land \text E)$

4) $\bot$ --- from 1) and 2), by $(\lnot \text E)$ (alternatively, using $(\to \text E)$, if $\lnot P$ is defined as $P \to \bot$)

5) $\lnot (P \land Q)$ --- from 2) and 4) by $(\lnot \text I)$, discharging [a].


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