1. ⇒ (MHT CET 2023 11th May Evening Shift )

The logical statement $(\sim (\sim p \lor q) \lor (p \land r)) \land (\sim q \land r)$ is equivalent to

A. $\sim p \lor r$

B. $(p \land \sim q) \lor r$

C. $(p \land r) \land \sim q$

D. $(\sim p \land \sim q) \land r$

Correct answer option is (C)

$\begin{aligned} [\sim (\sim p \lor q) \lor (p \land r)] \land (\sim q \land r) \\ \equiv [(p \land \sim q) \lor (p \land r)] \land (\sim q \land r) ...[De Morgan's law] \end{aligned}$

$\begin{aligned} \equiv p \land (\sim q \lor r) \land (\sim q \land r) \dots [Distributive law] \\ \equiv p \land [(\sim q \lor r) \land \sim q] \land r \dots A Associative law] \\ \equiv p \land (\sim q) \land r \dots [Absorption law] \\ \equiv (p \land r) \land \sim q ...[Commutative and Associative law] \end{aligned}$

2. ⇒ (MHT CET 2023 11th May Morning Shift )

The statement pattern $p \to\sim (p \land \sim q)$ is equivalent to

A. $q$

B. $(\sim p) \lor q$

C. $(\sim p) \land q$

D. $(\sim p) \lor (\sim q)$

Correct answer option is (B)

$\begin{aligned} p \to\sim (p \land \sim q) \\ \equiv\sim p \lor \sim (p \land \sim q) \dots [∵ p \to q \equiv\sim p \lor q] \\ \equiv\sim p \lor (\sim p \lor q) ...[De Morgan's law] \\ \equiv (\sim p \lor \sim p) \lor q ...[Associative law] \\ \equiv\sim p \lor q ...[Idempotent law] \end{aligned}$

3. ⇒ (MHT CET 2023 10th May Evening Shift )

The negation of the statement pattern $\sim s \lor (\sim r \land s)$ is equivalent to

A. $s \land r$

B. $s \land (r \land \sim s)$

C. $s \land \sim r$

D. $s \lor (r \lor \sim s)$

Correct answer option is (A)

$\begin{aligned} \sim (\sim s \lor (\sim r \land s)) \\ \equiv s \land \sim (\sim r \land s) . . . [De Morgan's law] \\ \equiv s \land (r \lor \sim s) . . . [De Morgan's law] \\ \equiv (s \land r) \lor (s \land \sim s) . . . [Distributive law] \\ \equiv (s \land r) \lor F . . . [Complement law] \\ \equiv s \land r . . . [Identity law] \end{aligned}$

4. ⇒ (MHT CET 2023 10th May Morning Shift )

The logical statement $[\sim (\sim p \lor q) \lor (p \land r)] \land (\sim q \land r)$ is equivalent to

A. $(p \land r) \land \sim q$

B. $(p \land \sim q) \lor r$

C. $\sim p \lor r$

D. $\sim p \land r$

Correct answer option is (A)

$[\sim (\sim p \lor q) \lor (p \land r)] \land (\sim q \land r)$

$\equiv [(p \land \sim q) \lor (p \land r)] \land (\sim q \land r)$ ...[De Morgan's law]

$\equiv p \land (\sim q \lor r) \land (\sim q \land r)$ ...[Distributive law]

$\equiv p \land [(\sim q \lor r) \land \sim q] \land r$ ...[Associative law]

$\equiv p \land (\sim q) \land r$ ....[Absorption law]

$\equiv (p \land r) \land \sim q$ ...[Commutative law]

5. ⇒ (MHT CET 2021 21th September Evening Shift )

The logical statement (p $\to$ q) $\land$ (q $\to$ ~p) is equivalent to

A. ~p

B. p

C. q

D. ~q

Correct answer option is (A)

$\begin{aligned} (p \to q) \land (q \to\sim p) \\ \equiv (\sim p \lor q) \land (\sim q \lor \sim p) \\ = (\sim p) \lor (q \land \sim q) \\ \equiv (\sim p) \lor F \equiv\sim p \end{aligned}$

⇒ Mathematical Logic