Disjunctive simplification

Question

What is this rule of inference called?

$(P\wedge Q)\vee(P\wedge\neg Q)\vdash P$

My (silly) motivation is this answer.

Answer 1

It can be justified by distribution and the law of the excluded middle: $Q\lor \lnot Q \equiv T$:

$$(P \land Q)\lor (P \land \lnot Q) \equiv P \land (Q\lor \lnot Q) \equiv P$$

Answer 2

You don't need to use the law of excluded middle (i.e. this is true in intuitionistic logic too). In fact even a stronger statement is true:

$$(P \land A) \lor (P \land B) \to P$$

and you can prove it by the distributive law and projection (i.e. $P \land Q \to P$):

\begin{align} (P \land A) \lor (P \land B) &\to P\\ P \land (A \lor B) &\to P \\ P &\to P \end{align}

I hope this helps $\ddot\smile$

Answer 3

Someone down voted dtldarek's answer, but his principal claim is plainly correct:

In a natural deduction system, start with the premiss:

$(P \land A) \lor (P \land B)$

then argue by cases (an intuitionistically acceptable mode of reasoning):

$\quad|\quad (P \land A)$

$\quad|\quad P$

$\quad/$

$\quad|\quad (P \land B)$

$\quad|\quad P$

Since we get to the same conclusion either way, we can discharge the two temporary assumptions and conclude

$P$.

Hence we have $(P\wedge A)\vee(P\wedge B)\vdash P$ whatever the $A$ and $B$, and so have $(P\wedge Q)\vee(P\wedge\neg Q)\vdash P$ as a special case. The special case doesn't depend on the law of excluded middle.