MHT CET · Maths · Mathematical Reasoning
The logical statement
\((\sim(\sim \mathrm{p} \vee \mathrm{q}) \vee(\mathrm{p} \wedge \mathrm{r})) \wedge(\sim \mathrm{q} \wedge \mathrm{r})\) is equivalent to
- A \(\sim \mathrm{p} \vee \mathrm{r}\)
- B \((p \wedge \sim q) \vee r\)
- C \((\mathrm{p} \wedge \mathrm{r}) \wedge \sim \mathrm{q}\)
- D \((\sim p \wedge \sim q) \wedge r\)
Answer & Solution
Correct Answer
(C) \((\mathrm{p} \wedge \mathrm{r}) \wedge \sim \mathrm{q}\)
Step-by-step Solution
Detailed explanation
\( {[\sim(\sim p \vee q) \vee(p \wedge r)] \wedge(\sim q \wedge r)} \)
\( \equiv[(p \wedge \sim q) \vee(p \wedge r)] \wedge(\sim q \wedge r) \Rightarrow \)\(\ldots[\text { De Morgan's law }]\)
\(\equiv \mathrm{p} \wedge(\sim \mathrm{q} \vee \mathrm{r}) \wedge(\sim \mathrm{q} \wedge \mathrm{r}) \quad \ldots[\) Distributive law \(]\)
\(\equiv p \wedge[(\sim q \vee r) \wedge \sim q] \wedge r^{\prime} .. [\text { Associative law] }\)
\(\equiv \mathrm{p} \wedge(\sim \mathrm{q}) \wedge\mathrm{r} \quad \ldots[\) Absórption law]
\(\equiv(p \wedge r) \wedge \sim q\)
.. [Commutative and Associative law]
\( \equiv[(p \wedge \sim q) \vee(p \wedge r)] \wedge(\sim q \wedge r) \Rightarrow \)\(\ldots[\text { De Morgan's law }]\)
\(\equiv \mathrm{p} \wedge(\sim \mathrm{q} \vee \mathrm{r}) \wedge(\sim \mathrm{q} \wedge \mathrm{r}) \quad \ldots[\) Distributive law \(]\)
\(\equiv p \wedge[(\sim q \vee r) \wedge \sim q] \wedge r^{\prime} .. [\text { Associative law] }\)
\(\equiv \mathrm{p} \wedge(\sim \mathrm{q}) \wedge\mathrm{r} \quad \ldots[\) Absórption law]
\(\equiv(p \wedge r) \wedge \sim q\)
.. [Commutative and Associative law]
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