Hypothetical Syllogism/Formulation 3/Proof by Truth Table

Theorem

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

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}$