Rule of Exportation/Reverse Implication/Formulation 2/Proof by Truth Table
Theorem
- $\vdash \paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \land q} \implies r}$
Proof
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations.
$\begin{array}{|ccccc|c|ccccc|} \hline (p & \implies & (q & \implies & r)) & \implies & ((p & \land & q) & \implies & r) \\ \hline \F & \T & \F & \T & \F & \T & \F & \F & \F & \T & \F \\ \F & \T & \F & \T & \T & \T & \F & \F & \F & \T & \T \\ \F & \T & \T & \F & \F & \T & \F & \F & \T & \T & \F \\ \F & \T & \T & \T & \T & \T & \F & \F & \T & \T & \T \\ \T & \T & \F & \T & \F & \T & \T & \F & \F & \T & \F \\ \T & \T & \F & \T & \T & \T & \T & \F & \F & \T & \T \\ \T & \F & \T & \F & \F & \T & \T & \T & \T & \F & \F \\ \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T \\ \hline \end{array}$
$\blacksquare$
Sources
- 1946: Alfred Tarski: Introduction to Logic and to the Methodology of Deductive Sciences (2nd ed.) ... (previous) ... (next): $\S \text{II}$: Exercise $13$