Destructive Dilemma/Formulation 1

Definition

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

Its abbreviation in a tableau proof is $\textrm{DD}$.

Proof by Natural Deduction
By the tableau method:

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.