MHT CET · Maths · Mathematical Reasoning
The statement pattern \(\mathrm{p} \wedge(\mathrm{q} \vee \sim \mathrm{p})\) is equivalent to
- A \(\mathrm{p} \wedge \mathrm{q}\)
- B \(\mathrm{p} \rightarrow \mathrm{q}\)
- C \(\mathrm{q} \wedge \sim \mathrm{p}\)
- D \(\mathrm{p} \vee \mathrm{q}\)
Answer & Solution
Correct Answer
(A) \(\mathrm{p} \wedge \mathrm{q}\)
Step-by-step Solution
Detailed explanation
(C)
\(\begin{array}{ll}\mathrm{p} \wedge(\mathrm{q} \vee \sim \mathrm{p}) & \\ \equiv(\mathrm{p} \wedge \mathrm{q}) \vee(\mathrm{p} \wedge \sim \mathrm{p}) & {[\text { Distributive law }]} \\ \equiv(\mathrm{p} \wedge \mathrm{q}) \vee \mathrm{F} & {[\text { complement law }]} \\ \equiv \mathrm{p} \wedge \mathrm{q} & {[\text { Identity law }]}\end{array}\)
\(\begin{array}{ll}\mathrm{p} \wedge(\mathrm{q} \vee \sim \mathrm{p}) & \\ \equiv(\mathrm{p} \wedge \mathrm{q}) \vee(\mathrm{p} \wedge \sim \mathrm{p}) & {[\text { Distributive law }]} \\ \equiv(\mathrm{p} \wedge \mathrm{q}) \vee \mathrm{F} & {[\text { complement law }]} \\ \equiv \mathrm{p} \wedge \mathrm{q} & {[\text { Identity law }]}\end{array}\)
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