Question

The statement $(p\vee \sim q)\wedge (\sim p\vee q)$ is

• A. a contradiction
• B. a tautology
• C. logically equivalent to $p\wedge q$
• D. neither a contradiction nor a tautology

Answer 1

The correct option is D neither a contradiction nor a tautology

The truth table of given expression is given below
$\begin{array}{ccccccc}p& q& \sim p& \sim q& x\equiv (p\vee \sim q)& y\equiv (\sim p\vee q)& x\wedge y\\ T& T& F& F& T& T& T\\ T& F& F& T& T& F& F\\ F& T& T& F& F& T& F\\ F& F& T& T& T& T& T\end{array}$

Clearly, the truth value of given preposition is sometimes true and sometimes false, hence $(p\vee \sim q)\wedge (\sim p\vee q)$ is neither a contradiction nor a tautology.