Factor Principles/Conjunction on Left/Formulation 1/Proof 1

Theorem

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

Proof

By the tableau method of natural deduction:

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

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$p \implies q$	Premise	(None)
2			$r \implies r$	Law of Identity	(None)	This is a theorem so depends on nothing
3		1	$\paren {r \implies r} \land \paren {p \implies q}$	Rule of Conjunction: $\land \II$	2, 1
4		1	$\paren {r \land p} \implies \paren {r \land q}$	Sequent Introduction	3	Praeclarum Theorema

$\blacksquare$