For integers $m$ and $n$, both greater than 1, consider the following three statements : $P : m$ divides $n$, $Q : m$ divides $n ^ { 2 }$, $R : m$ is prime, then
(1) $Q \wedge R \rightarrow P$
(2) $P \wedge Q \rightarrow R$
(3) $Q \rightarrow R$
(4) $Q \rightarrow P$

The Boolean expression $( p \wedge \sim q ) \Rightarrow ( q \vee \sim p )$ is equivalent to:
(1) $q \Rightarrow p$
(2) $p \Rightarrow q$
(3) $\sim q \Rightarrow p$
(4) $p \Rightarrow \sim q$

The negation of the Boolean expression $(\sim q \wedge p) \Rightarrow (\sim p \vee q)$ is logically equivalent to
(1) $p \Rightarrow q$
(2) $q \Rightarrow p$
(3) $\sim p \Rightarrow q$
(4) $\sim q \Rightarrow p$

For $\alpha \in \mathbb{N}$, consider a relation $R$ on $\mathbb{N}$ given by $R = \{(x, y) : 3x + \alpha y \text{ is a multiple of } 7\}$. The relation $R$ is an equivalence relation if and only if
(1) $\alpha = 14$
(2) $\alpha$ is a multiple of 4
(3) 4 is the remainder when $\alpha$ is divided by 10
(4) 4 is the remainder when $\alpha$ is divided by 7

The statement $( p \wedge q ) \Rightarrow ( p \wedge r )$ is equivalent to
(1) $q \Rightarrow ( p \wedge r )$
(2) $p \Rightarrow ( p \wedge r )$
(3) $( p \wedge r ) \Rightarrow ( p \wedge q )$
(4) $( p \wedge q ) \Rightarrow r$

The statement $(p \Rightarrow (q \vee p)) \Rightarrow r$ is NOT equivalent to:
(1) $p \wedge \sim r \Rightarrow q$
(2) $\sim q \Rightarrow (\sim r \vee p)$
(3) $p \Rightarrow (q \vee r)$
(4) $p \wedge \sim q \Rightarrow r$

The compound statement $( \sim ( P \wedge Q ) ) \vee ( ( \sim P ) \wedge Q ) \Rightarrow ( ( \sim P ) \wedge ( \sim Q ) )$ is equivalent to
(1) $( ( \sim P ) \vee Q ) \wedge ( ( \sim Q ) \vee P )$
(2) $( \sim Q ) \vee P$
(3) $( ( \sim P ) \vee Q ) \wedge ( \sim Q )$
(4) $( \sim P ) \vee Q$

Let $p$ and $q$ be two statements. Then $\sim ( p \wedge ( p \rightarrow \sim q ) )$ is equivalent to
(1) $p \vee ( p \wedge ( \sim q ) )$
(2) $p \vee ( ( \sim p ) \wedge q )$
(3) $( \sim p ) \vee q$
(4) $p \wedge q$

The negation of $( p \wedge ( - q ) ) \vee ( - p )$ is equivalent to
(1) $p \wedge ( - q )$
(2) $p \wedge q$
(3) $p \vee ( q \vee ( - p ) )$
(4) $p \wedge ( q \wedge ( - p ) )$

The statement $\sim p \vee ( \sim p \wedge q )$ is equivalent to
(1) $\sim p \wedge q$
(2) $p \wedge q \wedge \sim p$
(3) $\sim p \wedge q \wedge q$
(4) $\sim p \vee q$

The statement $( p \wedge ( \sim q ) ) \vee ( ( \sim p ) \wedge q ) \vee ( ( \sim p ) \wedge ( \sim q ) )$ is equivalent to
(1) $\sim p \vee q$
(2) $\sim p \vee \sim q$
(3) $p \vee \sim q$
(4) $p \vee q$

Let $\mathbf { p } , \mathbf { q }$ and $\mathbf { r }$ be propositions with their negations denoted by $\mathbf { p } ^ { \prime } , \mathbf { q } ^ { \prime } , \mathbf { r } ^ { \prime }$ respectively. Which of the following is equivalent to the proposition
$$p \vee q \Rightarrow q \wedge r$$
?
A) $\mathrm { p } ^ { \prime } \wedge \mathrm { q } ^ { \prime } \Rightarrow \mathrm { q } ^ { \prime } \vee \mathrm { r } ^ { \prime }$
B) $\mathrm { p } ^ { \prime } \wedge \mathrm { q } ^ { \prime } \Rightarrow \mathrm { q } ^ { \prime } \wedge \mathrm { r } ^ { \prime }$
C) $p ^ { \prime } \vee q ^ { \prime } \Rightarrow q ^ { \prime } \wedge r ^ { \prime }$
D) $q ^ { \prime } \wedge r ^ { \prime } \Rightarrow p ^ { \prime } \vee q ^ { \prime }$
E) $q ^ { \prime } \vee r ^ { \prime } \Rightarrow p ^ { \prime } \wedge q ^ { \prime }$