Destructive Dilemma/Formulation 1

Definition

 * $p \implies q, r \implies s \vdash \neg q \lor \neg s \implies \neg p \lor \neg r$

Proof

 * align="right" | 5 ||
 * align="right" | 1, 4
 * $\neg p$
 * MTT
 * 1, 4
 * 1, 4


 * align="right" | 8 ||
 * align="right" | 2, 7
 * $\neg r$
 * MTT
 * 2, 7
 * 2, 7


 * align="right" | 11 ||
 * align="right" | 1, 2
 * $\neg q \lor \neg s \implies \neg p \lor \neg r$
 * $\implies \mathcal I$
 * 3, 10
 * 3, 10

Alternative Proof

 * align="right" | 3 ||
 * align="right" | 1, 2
 * $\left({p \land r}\right) \implies \left({q \land s}\right)$
 * Sequent Introduction
 * 1, 2
 * Praeclarum Theorema
 * Praeclarum Theorema


 * align="right" | 5 ||
 * align="right" | 4
 * $\neg \left({q \land s}\right)$
 * Sequent Introduction
 * 4
 * De Morgan's Laws: Disjunction of Negations
 * align="right" | 6 ||
 * align="right" | 1, 2
 * $\neg \left({p \land r}\right)$
 * MTT
 * 3, 5
 * align="right" | 7 ||
 * align="right" | 1, 2
 * $\neg p \lor \neg r$
 * Sequent Introduction
 * 6
 * De Morgan's Laws: Disjunction of Negations
 * $\neg p \lor \neg r$
 * Sequent Introduction
 * 6
 * De Morgan's Laws: Disjunction of Negations

Proof by Truth Table
We apply the Method of Truth Tables to the proposition.

As can be seen for all models by inspection, where the truth value under the main connective on the LHS is $T$, that under the one on the RHS is also $T$:


 * $\begin{array}{|ccccccc||ccccccccccc|} \hline

(p & \implies & q) & \land & (r & \implies & s) & (\neg & q & \lor & \neg & s) & \implies & (\neg & p & \lor & \neg & r) \\ \hline F & T & F & T & F & T & F & T & F & T & T & F & T & T & F & T & T & F \\ F & T & F & T & F & T & T & T & F & T & F & T & T & T & F & T & T & F \\ F & T & F & F & T & F & F & T & F & T & T & F & T & T & F & T & F & T \\ F & T & F & T & T & T & T & T & F & T & F & T & T & T & F & T & F & T \\ F & T & T & T & F & T & F & F & T & T & T & F & T & T & F & T & T & F \\ F & T & T & T & F & T & T & F & T & F & F & T & T & T & F & T & T & F \\ F & T & T & F & T & F & F & F & T & T & T & F & T & T & F & T & F & T \\ F & T & T & T & T & T & T & F & T & F & F & T & T & T & F & T & F & T \\ T & F & F & F & F & T & F & T & F & T & T & F & T & F & T & T & T & F \\ T & F & F & F & F & T & T & T & F & T & F & T & T & F & T & T & T & F \\ T & F & F & F & T & F & F & T & F & T & T & F & F & F & T & F & F & T \\ T & F & F & F & T & T & T & T & F & T & F & T & F & F & T & F & F & T \\ T & T & T & T & F & T & F & F & T & T & T & F & T & F & T & T & T & F \\ T & T & T & T & F & T & T & F & T & F & F & T & T & F & T & T & T & F \\ T & T & T & F & T & F & F & F & T & T & T & F & F & F & T & F & F & T \\ T & T & T & T & T & T & T & F & T & F & F & T & T & F & T & F & F & T \\ \hline \end{array}$

Hence the result.

Note that the two formulas are not equivalent, as the relevant columns do not match exactly.