2
$\begingroup$

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)

$\endgroup$
3
$\begingroup$

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].

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.