Factor Principles/Conjunction on Right/Formulation 1/Proof 2

Theorem

$p \implies q \vdash \paren {p \land r} \implies \paren {q \land r}$

$p \implies q \vdash \paren {p \land r} \implies \paren {q \land r} $
Line	Pool	Formula	Rule	Depends upon	Notes
1	1	$p \implies q$	Premise	(None)
2	2	$p \land r$	Assumption	(None)
3	2	$p$	Rule of Simplification: $\land \EE_1$	2
4	1, 2	$q$	Modus Ponendo Ponens: $\implies \mathcal E$	1, 3
5	2	$r$	Rule of Simplification: $\land \EE_2$	2
6	1, 2	$q \land r$	Rule of Conjunction: $\land \II$	4, 5
7	1	$\paren {p \land r} \implies \paren {q \land r}$	Rule of Implication: $\implies \II$	2 – 6	Assumption 2 has been discharged

$\blacksquare$