MHT CET · Maths · Mathematical Reasoning
Negation of inverse of the following statement pattern \((p \wedge q) \rightarrow(p \vee \sim q)\) is
- A \(\mathrm{p}\)
- B \(\sim q\)
- C \(\sim p\)
- D \(\mathrm{q}\)
Answer & Solution
Correct Answer
(B) \(\sim q\)
Step-by-step Solution
Detailed explanation
Inverse of \((p \wedge q) \rightarrow(p \vee \sim q)\) is
\(
\begin{aligned}
& \sim(p \wedge q) \rightarrow \sim(p \vee \sim q) \\
& \equiv \sim[\sim(p \wedge q)] \vee \sim(p \vee \sim q) \ldots[p \rightarrow q \equiv \sim p \vee q]
\end{aligned}
\)
\(
\begin{aligned}
& \equiv(p \wedge q) \vee(\sim p \wedge q)...[De Morgan's law]\\
& \equiv(q \wedge p) \vee(q \wedge \sim p)...[Commutative law] \\
& \equiv q \wedge(p \vee \sim p)...[Distributive law] \\
& \equiv q \wedge T...[Complement law] \\
& \equiv q...[Identity law]
\end{aligned}
\)
\(\therefore \quad\) Negation of inverse of \((p \wedge q) \rightarrow(p \vee \sim q)\) is \(\sim q\)
\(
\begin{aligned}
& \sim(p \wedge q) \rightarrow \sim(p \vee \sim q) \\
& \equiv \sim[\sim(p \wedge q)] \vee \sim(p \vee \sim q) \ldots[p \rightarrow q \equiv \sim p \vee q]
\end{aligned}
\)
\(
\begin{aligned}
& \equiv(p \wedge q) \vee(\sim p \wedge q)...[De Morgan's law]\\
& \equiv(q \wedge p) \vee(q \wedge \sim p)...[Commutative law] \\
& \equiv q \wedge(p \vee \sim p)...[Distributive law] \\
& \equiv q \wedge T...[Complement law] \\
& \equiv q...[Identity law]
\end{aligned}
\)
\(\therefore \quad\) Negation of inverse of \((p \wedge q) \rightarrow(p \vee \sim q)\) is \(\sim q\)
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