Negation of the Boolean expression p Leftrigh | vedclass

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(D) We want to find the negation of the expression $p \Leftrightarrow (q \Rightarrow p)$.Let $S = p \Leftrightarrow (q \Rightarrow p)$.Using the prope

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$(\sim p) \wedge (\sim q)$

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(D) We want to find the negation of the expression $p \Leftrightarrow (q \Rightarrow p)$.Let $S = p \Leftrightarrow (q \Rightarrow p)$.Using the property $\sim(A \Leftrightarrow B) = (A \wedge \sim B) \vee (B \wedge \sim A)$,we have:$\sim S = (p \wedge \sim(q$ $\Rightarrow p)) \vee ((q$ $\Rightarrow p) \wedge \sim p)$.Now,simplify the first part: $p \wedge \sim(q \Rightarrow p) = p \wedge (q \wedge \sim p) = (p \wedge \sim p) \wedge q = F \wedge q = F$ (where $F$ is a contradiction).Simplify the second part: $(q$ $\Rightarrow p) \wedge \sim p = (\sim q \vee p) \wedge \sim p = (\sim q \wedge \sim p) \vee (p \wedge \sim p) = (\sim p \wedge \sim q) \vee F = \sim p \wedge \sim q$.Combining these,$\sim S = F \vee (\sim p \wedge \sim q) = \sim p \wedge \sim q$.

Negation of the Boolean expression $p \Leftrightarrow (q \Rightarrow p)$ is:

Similar Questions

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

Consider the following statement patterns:
$A$. $(q \to p) \vee (p \to q)$
$B$. $(\sim p \vee \sim q) \leftrightarrow \sim (p \wedge q)$
$C$. $[(p \vee q) \wedge \sim p] \wedge \sim q$
$D$. $(p \wedge q) \wedge (\sim p \vee \sim q)$
Which of the following is correct?

Statement-$I$: $(p \wedge \sim q) \wedge (\sim p \wedge q)$ is a contradiction.
Statement-$II$: $(p$ $\rightarrow q) \Leftrightarrow (\sim q$ $\rightarrow \sim p)$ is a tautology.

For the circuit shown below,the Boolean polynomial is

Write the negation of the following statement:
Both the diagonals of a rectangle have the same length.

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