If p, q, r are simple propositions,then (p we | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given logical expression is $(p \wedge q) \wedge (q \wedge r) = T$. For the conjunction of two statements to be true,both components must be t

Vedclass · Accepted Answer

$p, q, r$ are all true

Vedclass · Answer

(B) The given logical expression is $(p \wedge q) \wedge (q \wedge r) = T$. For the conjunction of two statements to be true,both components must be true. Therefore,$(p \wedge q) = T$ and $(q \wedge r) = T$. For $(p \wedge q) = T$,both $p$ and $q$ must be true. For $(q \wedge r) = T$,both $q$ and $r$ must be true. Since $q$ is true in both cases,and $p$ and $r$ must also be true,it follows that $p, q, r$ are all true.

If $p, q, r$ are simple propositions,then $(p \wedge q) \wedge (q \wedge r)$ is true,then:

Similar Questions

For the circuit shown below,the Boolean polynomial is

Write the negation of the following statement:
$\sqrt{2}$ is not a complex number.

Let $p$ and $q$ be two statements. Then $\sim(p \wedge (p \Rightarrow \sim q))$ is equivalent to:

$p :$ Suman is brilliant.
$q :$ Suman is rich.
$r :$ Suman is honest.
How can the negation of the statement "Suman is rich if and only if Suman is brilliant and dishonest" be represented?

The dual of $(x \vee y) \wedge (x \vee 1) = x \vee (x \wedge y) \vee y$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module