Self-Distributive Law for Conditional/Formulation 1/Proof by Truth Table

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Theorem

$p \implies \paren {q \implies r} \dashv \vdash \paren {p \implies q} \implies \paren {p \implies r}$


Proof

We apply the Method of Truth Tables to the proposition: $p \implies \paren {q \implies r} \dashv \vdash \paren {p \implies q} \implies \paren {p \implies r}$

As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.

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

$\blacksquare$

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