= Solution
Let $F=(A\to C)\lor(B\to C)$ and $H=(A\lor B)\to C$, so the two displayed <propositional formulas> are $\varphi=F\to H$ and $\psi=H\to F$. Their complete <truth tables> are
$$
\begin{array}{ccc|cc|cc|cc}
A&B&C&A\to C&B\to C&F&H&\varphi&\psi\\\hline
0&0&0&1&1&1&1&1&1\\
0&0&1&1&1&1&1&1&1\\
0&1&0&1&0&1&0&0&1\\
0&1&1&1&1&1&1&1&1\\
1&0&0&0&1&1&0&0&1\\
1&0&1&1&1&1&1&1&1\\
1&1&0&0&0&0&0&1&1\\
1&1&1&1&1&1&1&1&1
\end{array}
$$
Therefore \b[$\psi$ is a <tautology>; $\varphi$ is not]. For example $(A,B,C)=(0,1,0)$ makes the antecedent $F$ true via $A\to C$, while $H$ is false because $B$ is true and $C$ is false. This explains why having just one of the two <logical implications> is insufficient for the forward direction.
Back to article page
