1
$\begingroup$

If $p \lor q$ means that $p$ is true, $q$ is true, or both are true. Then why can $p \lor q \vdash r$ be proved by proving $p \vdash r$ and $q \vdash r$, without proving $p \land q \vdash r$?

Wouldn't this mean that $(p \lor q) \vdash \lnot (p \land q)$, which seems obviously false?

$\endgroup$
2
$\begingroup$

$p\lor q$ means that $p$ is true or $q$ is true. Note that if $p$ and $q$ are true, then $p$ is true, so that case is already covered. If you have proved that $p\vdash r$ and $q\vdash r$, then $p\lor q\vdash r$ automatically. This is, again, because the only way $p\lor q$ can be true, is if $p$ or $q$ is true. $p\lor q$ being true splits into two cases: (1) $p$ is true and (2) $q$ is true.

Saying this differently. $p\land q$ implies $p$ and $p\land q$ implies $q$. So there is not need to consider $p\land q$ as a separate case.

$\endgroup$
3
  • $\begingroup$ My problem is that $p \vdash \lnot(p \land q)$ and $q \vdash \lnot(p \land q)$ are provable, but $(p \land q) \vdash \lnot(p \land q)$ is not. $\endgroup$ – TheSandwichMakr Aug 4 '16 at 0:14
  • 2
    $\begingroup$ @TheSandwichMakr: How is $p\vdash \neg(p\wedge q)$ provable? $\endgroup$ – Eric Wofsey Aug 4 '16 at 0:19
  • $\begingroup$ @EricWofsey Oh, thanks, that clears everything up. $\endgroup$ – TheSandwichMakr Aug 4 '16 at 0:25

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.