Destructive Dilemma/Formulation 1/Proof 2

Theorem

$\ds p \implies q$

$\ds $

$\ds r \implies s$

$\ds $

$\ds \vdash \ \ $

$\ds \neg q \lor \neg s \implies \neg p \lor \neg r$

$\ds $

$p \implies q, r \implies s \vdash \neg q \lor \neg s \implies \neg p \lor \neg r$
Line	Pool	Formula	Rule	Depends upon	Notes
1	1	$p \implies q$	Premise	(None)
2	2	$r \implies s$	Premise	(None)
3	1, 2	$\paren {p \land r} \implies \paren {q \land s}$	Sequent Introduction	1, 2	Praeclarum Theorema
4	4	$\neg q \lor \neg s$	Assumption	(None)
5	4	$\neg \paren {q \land s}$	Sequent Introduction	4	De Morgan's Laws: Disjunction of Negations
6	1, 2, 4	$\neg \paren {p \land r}$	Modus Tollendo Tollens (MTT)	3, 5
7	1, 2	$\neg p \lor \neg r$	Sequent Introduction	6	De Morgan's Laws: Disjunction of Negations
8	1, 2	$\neg q \lor \neg s \implies \neg p \lor \neg r$	Rule of Implication: $\implies \II$	4 – 7	Assumption 4 has been discharged

$\blacksquare$