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?


$p\lor q$ 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.

  • $\begingroup$ 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

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