MHT CET · Maths · Mathematical Reasoning
Let \(\mathrm{p}, \mathrm{q}, \mathrm{r}\) be three statements, then \([\mathrm{p} \rightarrow(\mathrm{q} \rightarrow \mathrm{r})] \leftrightarrow[(\mathrm{p} \wedge \mathrm{q}) \rightarrow \mathrm{r}]\) is
- A equivalent to \(\mathrm{p} \leftrightarrow \mathrm{q}\).
- B contingency.
- C tautology.
- D contradiction.
Answer & Solution
Correct Answer
(C) tautology.
Step-by-step Solution
Detailed explanation
Given statement,
\({[p \rightarrow(q \rightarrow r)] } \leftrightarrow[(p \wedge q) \rightarrow r] \)
\( p \rightarrow(q \rightarrow r) \equiv \sim p \vee(q \rightarrow r) \)
\( \equiv \sim p \vee(\sim q \vee r) \)
\( \equiv[(\sim p) \vee(-q)] \vee r ... [\text { Associative law}]\)
\( \equiv \sim(p \wedge q) \vee r \quad \ldots[\text { De'morgans law }] \)
\( \equiv p \wedge q \rightarrow r\)
\(\therefore\)Given statement is tautology.
\({[p \rightarrow(q \rightarrow r)] } \leftrightarrow[(p \wedge q) \rightarrow r] \)
\( p \rightarrow(q \rightarrow r) \equiv \sim p \vee(q \rightarrow r) \)
\( \equiv \sim p \vee(\sim q \vee r) \)
\( \equiv[(\sim p) \vee(-q)] \vee r ... [\text { Associative law}]\)
\( \equiv \sim(p \wedge q) \vee r \quad \ldots[\text { De'morgans law }] \)
\( \equiv p \wedge q \rightarrow r\)
\(\therefore\)Given statement is tautology.
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