We verify b) and c) (De Morgan's laws) using a) (double-negation law).
a) $\lnot (\lnot P) \leftrightarrow P$.
b) - Start with the left-hand side and put $\lnot \lnot P$ in place of $P$ and $\lnot \lnot Q$ in place of $Q$ (i.e., use double-negation a)) :
$\lnot (P \lor Q) \leftrightarrow \lnot (\lnot \lnot P \lor \lnot \lnot Q)$
then use c) to transform the content of right-hand side parentheses into : $\lnot (\lnot P \land \lnot Q)$ [ rewrite it as : $\lnot [\lnot (\lnot P) \lor \lnot (\lnot Q) ]$ ; now it is of the "form" : $\lnot [\lnot P_1 \lor \lnot Q_1]$; then you must replace $\lnot P_1 \lor \lnot Q_1$ with $\lnot (P_1 \land Q_1)$, by c), that is really : $\lnot (\lnot P \land \lnot Q)$]. In this way you will get :
$\lnot (P \lor Q) \leftrightarrow \lnot (\lnot \lnot P \lor \lnot \lnot Q) \leftrightarrow \lnot \lnot (\lnot P \land \lnot Q)$
then apply again double-negation to the right-hand side ("cancelling" $\lnot \lnot$) and you will have :
$\lnot (P \lor Q) \leftrightarrow (\lnot P \land \lnot Q)$.
c) - Start with the left-hand side and put $\lnot \lnot P$ in place of $P$ and $\lnot \lnot Q$ in place of $Q$ (i.e., use double-negation a)) :
$\lnot (P \land Q) \leftrightarrow \lnot (\lnot \lnot P \land \lnot \lnot Q)$
then use b) to transform the content of right-hand side parentheses into : $\lnot (\lnot P \lor \lnot Q)$ getting :
$\lnot (P \land Q) \leftrightarrow \lnot (\lnot \lnot P \land \lnot \lnot Q) \leftrightarrow \lnot \lnot (\lnot P \lor \lnot Q)$
then apply again double-negation and it's done.
To prove a statement is a contingency you need to show that it is possible for the statement to be true, as well as show that it is possible for the statement to be false.
Now, it doesn't take much experience with logic to recognize $\neg (q \lor p \lor r)$ as a contingency, so for some audiences what you did will be enough. However, for a hard proof, you probably want to come up with two different truth-valuie assignment: one that sets the statement to True, and another on that sets the statement to false.
For example, to set the statement to false, we can set $p=q=r=True$, for then the statement evaluates to $\neg (q \lor p \lor r) =\neg (T \lor T \lor T) = \neg T = F$
Can you come up with a truth-value assignment that makes the statement True?
Best Answer
Use
XNORinstead ofequivalent.