Hypothetical Syllogism/Formulation 1/Proof by Truth Table

Theorem

 * $p \implies q, q \implies r \vdash p \implies r$

Proof
We apply the Method of Truth Tables to the propositions in turn.

As can be seen for all boolean interpretations by inspection, where the truth values under the main connective on the LHS is $T$, that under the one on the RHS is also $T$:

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

Hence the result.