The statement $(p \Rightarrow q) \vee(p \Rightarrow r)$ is NOT equivalent to.
$(p \wedge(\sim r)) \Rightarrow q$
$(\sim q) \Rightarrow((\sim r) \vee p)$
$p \Rightarrow(q \vee r)$
$(p \wedge(\sim q)) \Rightarrow r$
The negation of the compound proposition $p \vee (\sim p \vee q)$ is
Consider
Statement $-1 :$$\left( {p \wedge \sim q} \right) \wedge \left( { \sim p \wedge q} \right)$ is a fallacy.
Statement $-2 :$$(p \rightarrow q) \leftrightarrow ( \sim q \rightarrow \sim p )$ is a tautology.
The propositions $(p \Rightarrow \;\sim p) \wedge (\sim p \Rightarrow p)$ is a
The negation of the compound statement $^ \sim p \vee \left( {p \vee \left( {^ \sim q} \right)} \right)$ is
The statement $p \to ( q \to p)$ is equivalent to