The false statement in the following is \[\]
A. $p\wedge \tilde{\ }p$ is a contradiction\[\]
B.$\left( p\to q \right)\leftrightarrow \left( q\to p \right)$is a contradiction.\[\]
C. $\tilde{\ }\left( \tilde{\ }p \right)\leftrightarrow p$is a tautology \[\]
D. $p\vee \tilde{\ }\left( p \right)$is a tautology\[\]

Hint: We recall the basic operation of truth values of negation (Logical NOT ) conjunction (logical AND),disjunction (Logical OR) , implication (logical if else ) and bi-implication (logical if and only if ). We recall that a composite statement is called a tautology when it is true for all possible truth values of prime statements and their combination and is called a contradiction when it is false for all possible truth values of prime statements and their combination.\[\]

Complete step-by-step solution:
We know from the mathematical logic that if the statement $p$ has a truth value T or F then the negation of $p$ is denoted as $\tilde{\ }p$ and has truth value F or T respectively.
We also know that when there are two statements $p$ and $q$ , the statement with conjunction (with logical connective AND) of their truth values is denoted as $p\wedge q$ and has a truth value T only when both $p$ and $q$ have truth values T, otherwise false. The statement with disjunction (with logical connective OR) of their truth values is denoted as $p\hat{\ }q$ and has a truth value F only when both of $p$ and $q$ have truth value F, otherwise true.\[\]
The statement with implication (with logical connective If...then...) of their truth values is denoted as $p\to q$ and has a truth value F only when one of $p$ has a truth value T and $q$has a truth value $F$ otherwise true. The statement with bi-implication (with logical connective if and only if ) of their truth values is denoted as $p\leftrightarrow q$ and has a truth value T only when both $p$ and $q$ have truth value T or truth value F , otherwise F. \[\]
 We are give in the options four composite statements as follows$\left( A \right)p\wedge \tilde{\ }p,\left( B \right)\left( p\to q \right)\leftrightarrow \left( q\to p \right),\leftrightarrow \left( C \right)\tilde{\ }\left( \tilde{\ }p \right)p\left( D \right)p\vee \tilde{\ }\left( p \right)$. We see that in option (A) ,(B) and (D) single statements have been operated on. So we can solve it with one truth table. \[\]

We see that $p\wedge \tilde{\ }p$ is a contradiction since it's false for both truth values of $p$. The statement $\tilde{\ }\left( \tilde{\ }p \right)\to p$ is a tautology since its true for both truth values of and $p\vee \tilde{\ }\left( p \right)$ is similarly a tautology since So options A ,C, D are true statements. Let us draw truth table for option (B) $\left( p\to q \right)\leftrightarrow \left( q\to p \right)$. \[\]

We are given $\left( p\to q \right)\leftrightarrow \left( q\to p \right)$ is contradiction but since all truth values of the statement is not F it not a contradiction, Hence statement given in option B is false and the correct option is B. \[\]

Note: The statement $\tilde{\ }q\to \tilde{\ }p$ is the contra-positive of $p\to q$ and $q\to p$ is the converse of $p\to q$. We know that two composite statements are equivalent when they have the same truth value for all possible combinations of truth values for all prime statements. If we want to solve quickly we shall solve a statement involving one prime statement without writing and directly drawing the truth table for option B.