If $ p \rightarrow q $ and $q \rightarrow p$ are not tautolgies, is $ (p \rightarrow q) \rightarrow (q \rightarrow p)$ a tautology If found a multiple choice question online:
If $(p \rightarrow q) $ is not a tautology and $ (q \rightarrow p) $ is not a tautology, then:

*

*$ p \lor q $ is not a tautology

*$ p \lor q $ is a tautology

*$ p \land q $ is a contradiction

*$ (p \rightarrow q) \rightarrow (q \rightarrow p) $ is a tautology

*None of the above

I realized that for $ p \rightarrow q $ and $ q \rightarrow p $ to not be tautologies, there has to be a case where $ p $ is true and $ q $ is false, and vice versa. So I eliminated choices $ 1, 2, 3 $. But I don't understand what choice $ 4 $ means?
 A: 
But I don't understand what choice 4 means?

It means a counter example would have $p\to q$ be true but $q\to p$ be false.
The later may be false because $q\to p$ is not a tautology, and when it is false then $p\to q$ is ???
A: You could also find the answer if we took some logical reasoning using DeMorgan's Laws: We see that $(p \to q) $ is not a tautology, and $(q \to p)$ is also not a tautology, then:
$$(p \to q) \to (q \to p) \qquad \text{Given}
\\
\equiv (\lnot p \lor q) \to (\lnot q \lor p) \qquad p \to q \equiv \lnot p \lor q
\\
\equiv \lnot (\lnot p \lor q) \lor (\lnot q \lor p) \qquad p \to q \equiv \lnot p \lor q
\\
\equiv (p \land \lnot q) \lor (\lnot q \lor p) \qquad \text{Double Negation}
\\
\
\equiv \big[ p \lor (\lnot p \lor q) \big] \land \big[ q \lor (\lnot p \lor q) \big] \qquad \text{Distributive Law}
\\
\equiv [p \lor \lnot p \lor q] \land [\lnot p \lor q \lor q] \qquad \text{Associative Law}
\\
\equiv [T \lor q] \land [\lnot p \lor q] \qquad \text{Negation Law}
\\
\equiv T \land (\lnot p \lor q) \qquad \text{Domination Law}
\\
\equiv (\lnot p \lor q) \qquad \text{Identity Law}
\\
\equiv (p \to q) \qquad p \to q \equiv \lnot p \lor q
$$
We see that we end up with the given fact that we have defined that $p \to q$ is not a tautology. Since we started with $(p \to q) \to (q \to p)$ and ended with $p \to q$, we can conclude that this statement is indeed NOT  a tautology. $\square$
