Which of the following is false?

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(D) We evaluate the statement $((p \wedge p)$ $\rightarrow q)$ $\rightarrow p$. Since $p \wedge p \equiv p$,the expression simplifies to $(p$ $\righta

Vedclass · Accepted Answer

$((p \wedge p)$ $\rightarrow q)$ $\rightarrow p$ is a tautology.

Vedclass · Answer

(D) We evaluate the statement $((p \wedge p)$ $\rightarrow q)$ $\rightarrow p$. Since $p \wedge p \equiv p$,the expression simplifies to $(p$ $\rightarrow q)$ $\rightarrow p$. If $p$ is $F$ (False) and $q$ is $T$ (True) or $F$ (False),then $(p \rightarrow q)$ is $T$. Then $(p$ $\rightarrow q)$ $\rightarrow p$ becomes $T \rightarrow F$,which is $F$. Since the statement can be false,it is not a tautology. Therefore,option $D$ is false.

Similar Questions

Which of the following statement patterns is a tautology?
$S_1 \equiv (\sim q \wedge p) \wedge q$
$S_2 \equiv [p \wedge (p$ $\rightarrow q)]$ $\rightarrow q$
$S_3 \equiv (p \wedge q) \wedge (\sim p \vee \sim q)$
$S_4 \equiv (p \wedge q) \rightarrow r$

Among the statements:
$(S1): (p \Rightarrow q) \vee ((\sim p) \wedge q)$ is a tautology
$(S2): (q \Rightarrow p) \Rightarrow ((\sim p) \wedge q)$ is a contradiction

Simplify the Boolean function $(x \cdot y)+[(x+y') \cdot y]'$

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

The statement $p \to (q \to p)$ is equivalent to

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