Factor Principles/Disjunction on Right/Formulation 1/Proof 3

Theorem

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

Proof

By the tableau method of natural deduction:

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

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$p \implies q$	Premise	(None)
2		2	$p \lor r$	Assumption	(None)
3		2	$r \lor p$	Sequent Introduction	2	Disjunction is Commutative
4		1	$\paren {r \lor p} \implies \paren {r \lor q}$	Sequent Introduction	1	Factor Principles/Disjunction on Left/Formulation 1
5		1,2	$r \lor q$	Modus Ponendo Ponens: $\implies \mathcal E$	4, 3
6		1,2	$q \lor r$	Sequent Introduction	5	Disjunction is Commutative
7		1	$\paren {p \lor r} \implies \paren {q \lor r}$	Rule of Implication: $\implies \II$	2 – 6	Assumption 2 has been discharged

$\blacksquare$