MHT CET · Maths · Mathematical Reasoning
The statement pattern \([p \wedge(q \vee r)] \vee[\sim r \wedge \sim q \wedge p]\) is equivalent to
- A \(q \vee r\)
- B \(\mathrm{p} \vee \mathrm{r}\).
- C q
- D p
Answer & Solution
Correct Answer
(D) p
Step-by-step Solution
Detailed explanation
\( {[p \wedge(q \vee r)] \vee[\sim r \wedge \sim q \wedge p]} \)
\( \equiv[p \wedge(q \vee r)] \vee[p \wedge(\sim r \wedge \sim q)]\ldots \text{ [Commutativity}\) \(\text {and associativity]} \) \(\equiv[p \wedge(q \vee r)] \vee[p \wedge \sim(r \vee q)... \text{ [DeMorgan's Law]}\) \(\equiv p \wedge[(q \vee r) \vee \sim(q \vee r)] \ldots \text{...[Distributive and}\) \(\text{commutative Law]}\)
\( \equiv p \wedge T\text{...[Absorption Law]}\)
[Identity Law]
\( \equiv[p \wedge(q \vee r)] \vee[p \wedge(\sim r \wedge \sim q)]\ldots \text{ [Commutativity}\) \(\text {and associativity]} \) \(\equiv[p \wedge(q \vee r)] \vee[p \wedge \sim(r \vee q)... \text{ [DeMorgan's Law]}\) \(\equiv p \wedge[(q \vee r) \vee \sim(q \vee r)] \ldots \text{...[Distributive and}\) \(\text{commutative Law]}\)
\( \equiv p \wedge T\text{...[Absorption Law]}\)
[Identity Law]
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