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

Theorem

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

Proof
We apply the Method of Truth Tables to the proposition: $p \implies \left({q \implies r}\right) \dashv \vdash \left({p \implies q}\right) \implies \left({p \implies r}\right)$

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

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