The negation of the Boolean expression $p \vee (\sim p \wedge q)$ is equivalent to

The dual of $(x \vee y) \wedge (x \vee 1) = x \vee (x \wedge y) \vee y$ is

If $p: \forall n \in N, n^2+n$ is an even number and $q: \forall n \in N, n^2-n$ is an odd number,then the truth values of $p \wedge q, p \vee q$ and $p \rightarrow q$ are respectively:

Negation of $p \wedge (q \wedge \sim(p \wedge q))$ is

Simplify the following switching circuit and find the corresponding Boolean expression.

Let $p, q, r$ be three logical statements. Consider the compound statements $S_{1}: ((\sim p) \vee q) \vee ((\sim p) \vee r)$ and $S_{2}: p \rightarrow (q \vee r)$. Then,which of the following is $NOT$ true?