$(p \wedge r) \Leftrightarrow (p \wedge (\sim q))$ is equivalent to $(\sim p)$ when $r$ is.

The negation of the statement "If a quadrilateral is a square,then it is a rhombus" is:

If $p :$ It is raining today.
$q :$ $I$ go to school.
$r :$ $I$ will meet my friends.
$s :$ $I$ will go to watch a movie.
Then write the statement: 'If it does not rain today or $I$ do not go to school,then $I$ will meet my friends and go to watch a movie' in symbolic form.

The correct logical equivalences from the following are:
$(I)$ $p \to (q \to r) \equiv (p \land q) \to r$
$(II)$ $(p \to q) \to r \equiv p \to (q \lor r)$
$(III)$ $(p \to q) \to r \equiv (p \to r) \land (\sim q \to r)$
$(IV)$ $p \to (q \to r) \equiv q \to (p \to r)$

$(\sim (\sim p)) \wedge q$ is equal to .........

If $p$ : the function $f(x) = x$ is an even function,$q$ : ${}^nC_m$ is a whole number wherever defined,and $r$ : if $\overrightarrow{c} = \overrightarrow{a} + \lambda \overrightarrow{b}$,then the vectors $\overrightarrow{a}, \overrightarrow{b}$,and $\overrightarrow{c}$ are linearly dependent. Which one of the following statements is true?