Option A. $p \rightarrow (q \lor r) \equiv ((p \land \neg{q}) \rightarrow r) $
LHS:
$p \rightarrow (q \lor r)$ as p → q is same as $\neg{p} \lor q$
$ \neg{p} \lor q \lor r \\ $
$(p \land \neg{q}) \rightarrow r \\$ which is equivalent to RHS: $((p \land \neg{q}) \rightarrow r) $
Option B. $(p \land q) \lor r \not\equiv p \land (q \lor r)$
Option C. If $p \rightarrow q$ is False then p is True and q is False, then $q \rightarrow p$ is True as $False \rightarrow True$ is True
Option D. If $p \rightarrow q$ is True, then there 3 cases:
- p is True, q is True
- p is False, q is True
- p is False, q is False
then for case 1, 3 $q \rightarrow p$ is True which contracts the statements saying it is False.
Therefore B and D are right options i.e they are false statements