Rule of Exportation/Forward Implication/Formulation 2/Proof 2

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Theorem

$\vdash \left({\left ({p \land q}\right) \implies r}\right) \implies \left({p \implies \left ({q \implies r}\right)}\right)$


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.

$\begin{array}{|ccccc|c|ccccc|} \hline ((p & \land & q) & \implies & r) & \implies & (p & \implies & (q & \implies & r)) \\ \hline F & F & F & T & F & T & F & T & F & T & F \\ F & F & F & T & T & T & F & T & F & T & T \\ F & F & T & T & F & T & F & T & T & F & F \\ F & F & T & T & T & T & F & T & T & T & T \\ T & F & F & T & F & T & T & T & F & T & F \\ T & F & F & T & T & T & T & T & F & T & T \\ T & T & T & F & F & T & T & F & T & F & F \\ T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$

$\blacksquare$