Praeclarum Theorema/Formulation 1/Proof 1

Theorem

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

Proof

By the tableau method of natural deduction:

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

Line		Pool	Formula	Rule	Depends upon	Notes
1		1	$\paren {p \implies q} \land \paren {r \implies s}$	Premise	(None)
2		1	$p \implies q$	Rule of Simplification: $\land \EE_1$	1
3		1	$r \implies s$	Rule of Simplification: $\land \EE_2$	1
4		4	$p \land r$	Assumption	(None)
5		4	$p$	Rule of Simplification: $\land \EE_1$	4
6		4	$r$	Rule of Simplification: $\land \EE_2$	4
7		1, 4	$q$	Modus Ponendo Ponens: $\implies \mathcal E$	2, 5
8		1, 4	$s$	Modus Ponendo Ponens: $\implies \mathcal E$	3, 6
9		1, 4	$q \land s$	Rule of Conjunction: $\land \II$	7, 8
10		1	$\paren {p \land r} \implies \paren {q \land s}$	Rule of Implication: $\implies \II$	4 – 9	Assumption 4 has been discharged

$\blacksquare$