Conditional is Left Distributive over Conjunction/Formulation 2/Proof by Truth Table
Theorem
- $\vdash \paren {p \implies \paren {q \land r} } \iff \paren {\paren {p \implies q} \land \paren {p \implies r} }$
Proof
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations.
$\begin{array}{|ccccc|c|ccccccc|} \hline (p & \implies & (q & \land & r)) & \iff & ((p & \implies & q) & \land & (p & \implies & r)) \\ \hline \F & \T & \F & \F & \F & \T & \F & \T & \F & \T & \F & \T & \F \\ \F & \T & \F & \F & \T & \T & \F & \T & \F & \T & \F & \T & \T \\ \F & \T & \T & \F & \F & \T & \F & \T & \T & \T & \F & \T & \F \\ \F & \T & \T & \T & \T & \T & \F & \T & \T & \T & \F & \T & \T \\ \T & \F & \F & \F & \F & \T & \T & \F & \F & \F & \T & \F & \F \\ \T & \F & \F & \F & \T & \T & \T & \F & \F & \F & \T & \T & \T \\ \T & \F & \T & \F & \F & \T & \T & \T & \T & \F & \T & \F & \F \\ \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T \\ \hline \end{array}$
$\blacksquare$