Conditional is not Associative

Theorem

There exist propositions $p, q, r$ such that:

$p \implies \paren {q \implies r} \not \vdash \paren {p \implies q} \implies r$

That is, the conditional is not associative.

Proof

We apply the Method of Truth Tables:

$\begin{array}{|ccccc||ccccc|} \hline

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

As can be seen by inspection, the truth values under the main connectives do not match for all boolean interpretations.

$\blacksquare$