The contrapositive of the statement pattern [ | vedclass

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(D) The contrapositive of a conditional statement $P 	o Q$ is $\sim Q 	o \sim P$.Here, $P = [p \lor (p 	o q)]$ and $Q = (p \land \sim q)$.First, fi

Vedclass · Accepted Answer

$( \sim p \lor q) 	o [ \sim p \lor ( \sim p \lor q)]$

Vedclass · Answer

(D) The contrapositive of a conditional statement $P 	o Q$ is $\sim Q 	o \sim P$.Here, $P = [p \lor (p 	o q)]$ and $Q = (p \land \sim q)$.First, find $\sim Q$: $\sim (p \land \sim q) = \sim p \lor \sim (\sim q) = \sim p \lor q$.Next, find $\sim P$: $\sim [p \lor (p 	o q)] = \sim p \land \sim (p 	o q) = \sim p \land \sim (\sim p \lor q) = \sim p \land (p \land \sim q)$.Thus, the contrapositive $\sim Q 	o \sim P$ is $(\sim p \lor q) 	o [\sim p \land (p \land \sim q)]$.Comparing this with the options, option $(D)$ is the correct form.

The contrapositive of the statement pattern $[p \lor (p \to q)] \to (p \land \sim q)$ is

Similar Questions

Consider the following statements:
$r$: If $p \to q$ is false then $p \lor q$ is false.
$s$: If $p \leftrightarrow q$ is false then $p \lor q$ is false.
The truth values of $r \to s$ and $s \to r$ are respectively . . . . . .

The output of the following circuit is

The statement $p$ $\Rightarrow (q$ $\Rightarrow p)$ is logically equivalent to .....

$p \Rightarrow q$ can also be written as

Given the following two statements:
$(S_{1}): (q \vee p) \rightarrow (p \leftrightarrow \sim q)$ is a tautology.
$(S_{2}): \sim q \wedge (\sim p \leftrightarrow q)$ is a fallacy.
Then:

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