The negation of $(p \wedge (\sim q)) \vee (\sim p)$ is equivalent to:

The truth values of $p \rightarrow r$ is $F$ and $p \leftrightarrow q$ is $F$. Then the truth values of $(\sim p \vee q) \rightarrow (p \vee \sim q)$ and $(p \wedge \sim q) \rightarrow (\sim p \wedge q)$ are respectively:

The converse of the statement $((\sim p) \wedge q) \Rightarrow r$ is

Let $p, q, r$ denote arbitrary statements. Then the logical equivalent of the statement $p \Rightarrow (q \vee r)$ is

The maximum number of compound propositions,out of $p \vee r \vee s$,$p \vee \sim r \vee \sim s$,$p \vee \sim q \vee s$,$\sim p \vee \sim r \vee s$,$\sim p \vee \sim r \vee \sim s$,$\sim p \vee q \vee \sim s$,$q \vee r \vee \sim s$,$q \vee \sim r \vee \sim s$,$\sim p \vee \sim q \vee \sim s$ that can be made simultaneously true by an assignment of the truth values to $p, q, r$ and $s$,is equal to

The negation of the statement "$2 + 3 = 5$ and $8 < 10$" is: