(p to q) leftrightarrow (q vee sim p) is | vedclass

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(B) To determine the nature of the statement $(p 	o q) \leftrightarrow (q \vee \sim p)$,we construct a truth table:      $p$    $q$    $\sim p$    $p

Vedclass · Accepted Answer

Tautology

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(B) To determine the nature of the statement $(p 	o q) \leftrightarrow (q \vee \sim p)$,we construct a truth table:      $p$    $q$    $\sim p$    $p 	o q$    $q \vee \sim p$    $(p 	o q) \leftrightarrow (q \vee \sim p)$        $T$    $T$    $F$    $T$    $T$    $T$        $T$    $F$    $F$    $F$    $F$    $T$        $F$    $T$    $T$    $T$    $T$    $T$        $F$    $F$    $T$    $T$    $T$    $T$  Since the final column contains only $T$ (True) values,the statement is a tautology.

$(p \to q) \leftrightarrow (q \vee \sim p)$ is

Similar Questions

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

Are the following pairs of statements negations of each other:
$1$. The number $x$ is not a rational number.
$2$. The number $x$ is not an irrational number.

Statement-$I$: $\sim (p \leftrightarrow q)$ is equivalent to $(p \wedge \sim q) \vee (q \wedge \sim p)$.
Statement-$II$: $p$ $\rightarrow (p$ $\rightarrow q)$ is a tautology.

For the statement: "If a quadrilateral $ABCD$ is a rhombus,then its opposite sides are parallel",its contrapositive and converse are respectively given by:

If $(p \wedge \sim q) \wedge (p \wedge r) \rightarrow \sim p \vee q$ is false,then the truth values of $p, q$ and $r$ are respectively

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