Factor Principles/Conjunction on Left/Formulation 1

Theorem

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

Proof 1

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

$\blacksquare$

Proof 2

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

$\blacksquare$

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Sources

• 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $3$ Conjunction and Disjunction: Theorem $18$