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

Theorem

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

Proof

By the tableau method of natural deduction:

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

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$p \implies q$	Premise	(None)
2			$r \implies r$	Theorem Introduction	(None)	Law of Identity: Formulation 2
3		1	$\paren {r \lor p} \implies \paren {r \lor q}$	Sequent Introduction	2, 1	Constructive Dilemma

$\blacksquare$