Hypothetical Syllogism/Formulation 3/Proof 2

Theorem

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

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.

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

$\blacksquare$