Disjunctive simplification What is this rule of inference called?
$(P\wedge Q)\vee(P\wedge\neg Q)\vdash P$
My (silly) motivation is this answer.
 A: 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.
A: 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 $\ddot\smile$
A: It can be justified by distribution and the law of the excluded middle: $Q\lor \lnot Q \equiv T$: 
$$(P \land Q)\lor (P \land \lnot Q) \equiv P \land (Q\lor \lnot Q) \equiv  P$$
