Question

Which of the following statements is/are true? \[\]
A.$p\wedge \left( \tilde{\ }p \right)$ is a contradiction. \[\]
B.$\left( p\to q \right)\Leftrightarrow \left( \tilde{\ }q\to \tilde{\ }p \right)$ is contradiction. \[\]
C. $\tilde{\ }\left( \tilde{\ }p \right)\Leftrightarrow p$ is a tautology. \[\]
D. $p\vee \left( \tilde{\ }p \right)$ is a tautology. \[\]

Answer 1

Hint: We recall that a composite statement is called a tautology when the composite statement is true for all possible truth values combinations of prime statements and a contradiction when the composite statement is false for all possible truth values combinations of prime statements. We draw a truth table for each composite statement in the options and see if it is a tautology or contradiction or neither. \[\]

Complete step by step answer:
We know that we form composite statements from prime statements having truth values true (T) or false (F). We know that we call the composite statement $poq$ where $o$ is any logical operation a tautology when $poq$ is true for all truth values of $p,q$ and we call $poq$ a contradiction when
Let us check whether option A is true or not. Let us draw the truth table for $p\wedge \left( \tilde{\ }p \right)$. \[\]

$p$	$\tilde{\ }p$	$p\wedge \left( \tilde{\ }p \right)$
T	F	F
F	T	F

We see that the composite $p\wedge \left( \tilde{\ }p \right)$ is false for all possible truth values of $p$ and hence it is true that$p\wedge \left( \tilde{\ }p \right)$ is a contradiction. Let us check the option B and draw the truth table for $\left( p\to q \right)\Leftrightarrow \left( \tilde{\ }q\to \tilde{\ }p \right)$. \[\]

$p$	$q$	$\tilde{\ }p$	$\tilde{\ }q$	$p\to q$	$\tilde{\ }q\to \tilde{\ }p$	$\left( p\to q \right)\Leftrightarrow \left( \tilde{\ }q\to \tilde{\ }p \right)$
T	T	F	F	T	T	T
T	F	F	T	F	F	T
F	T	T	F	T	T	T
F	F	T	T	T	T	T

We see that the composite $\left( p\to q \right)\Leftrightarrow \left( \tilde{\ }q\to \tilde{\ }p \right)$ is true for all possible truth values of $p,q$ and hence it is false that$\left( p\to q \right)\Leftrightarrow \left( \tilde{\ }q\to \tilde{\ }p \right)$ is a contradiction because it is tautology. Let us draw truth table of $\tilde{\ }\left( \tilde{\ }p \right)\Leftrightarrow p$ to check option C. \[\]

$p$	$\tilde{\ }p$	$\tilde{\ }\left( \tilde{\ }p \right)$	$\tilde{\ }\left( \tilde{\ }p \right)\Leftrightarrow p$
T	F	T	F
F	T	F	F

We see that the composite $p\wedge \left( \tilde{\ }p \right)$ is false for all possible truth values of $p$ and hence it is true that$p\wedge \left( \tilde{\ }p \right)$ is a contradiction. Let us draw truth table of $p\vee \left( \tilde{\ }p \right)$ to check option D. \[\]

$p$	$\tilde{\ }p$	$p\vee \left( \tilde{\ }p \right)$
T	F	T
F	T	T

We see that the composite $p\vee \left( \tilde{\ }p \right)$ is true for all possible truth values of $p,q$ and hence it is true that $p\vee \left( \tilde{\ }p \right)$ is a tautology. So the correct options are A, C and D. \[\]

Note:
 We should always remember that We know 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. If there are two statements $p$ and $q$, the statement with conjunction $p\wedge q$ (with logical connective AND) has a truth value T only when both $p$ and $q$ have truth values T, otherwise F. The statement with disjunction $p\hat{\ }q$ (with logical connective OR) has a truth value F only when one of $p$ and $q$ have truth value F, otherwise T. The statement with implication $p\to q$ (with logical connective If...then...) has a truth value F only when one of $p$ has a truth value T and $q$has a truth value $F$ otherwise F. The statement with bi-implication $p\leftrightarrow q$ (with logical connective if and only if ) has a truth value T only when both $p$ and $q$ have truth value T or truth value F , otherwise F. The statement $\tilde{\ }p\to \tilde{\ }p$ is called contra-positive of $p\to q$.