Conjunction therefore Disjunction of Conjunctions with Complements

Theorem

$p \land q \vdash \left({p \land r}\right) \lor \left({q \land \neg r}\right)$

Proof

Proof by Natural deduction

This is proved by the Tableau method:

$p \land q \vdash \left({p \land r}\right) \lor \left({q \land \neg r}\right)$

Line		Pool	Formula	Rule	Depends upon
1		1	$p \land q$	P	(None)
2		-	$r \lor \neg r$	LEM	(None)
3		1	$p$	$\land \mathcal E_1$	1
4		1	$q$	$\land \mathcal E_2$	1
5		5	$r$	A	(None)
6		1, 5	$p \land r$	$\land \mathcal I$	1, 5
7		1, 5	$\left({p \land r}\right) \lor \left({q \land \neg r}\right)$	$\lor \mathcal I_1$	6
8		8	$\neg r$	A	(None)
9		1, 8	$q \land \neg r$	$\land \mathcal I$	4, 8
10		1, 8	$\left({p \land r}\right) \lor \left({q \land \neg r}\right)$	$\lor \mathcal I_2$	9
11		1	$\left({p \land r}\right) \lor \left(q \land \neg r\right)$	$\lor \mathcal E$	2, 5-7, 8-10	The assumptions at 5 and 8 are discharged.

$\blacksquare$