Factor Principles/Conjunction on Right/Formulation 1

Theorem

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

Proof 1

By the tableau method of natural deduction:

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

$\blacksquare$

Proof 2

By the tableau method of natural deduction:

$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

$\blacksquare$

Proof by Truth Table

Proof by Truth Table:

$\begin{array}{|ccc||ccccccccccc|} \hline p & q & r & (p & \implies & q) & \implies & ((p & \land & r) & \implies & (q & \land & r)) \\ \hline \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T \\ \T & \T & \F & \T & \T & \T & \T & \T & \F & \F & \T & \T & \F & \F \\ \T & \F & \T & \T & \F & \F & \T & \T & \T & \T & \F & \F & \F & \T \\ \T & \F & \F & \T & \F & \F & \T & \T & \F & \F & \T & \F & \F & \F \\ \F & \T & \T & \F & \T & \T & \T & \F & \F & \T & \T & \T & \T & \T \\ \F & \T & \F & \F & \T & \T & \T & \F & \F & \F & \T & \T & \F & \F \\ \F & \F & \T & \F & \T & \F & \T & \F & \F & \T & \T & \F & \F & \T \\ \F & \F & \F & \F & \T & \F & \T & \F & \F & \F & \T & \F & \F & \F \\ \hline \end{array}$

$\blacksquare$