Let \(p\) and \(q\) be any two propositions.
Statement 1 : \((p \rightarrow q) \leftrightarrow q \vee \sim p\) is a tautology.
Statement 2 : \(\sim(\sim p \wedge q) \wedge(p \vee q) \leftrightarrow p\) is a fallacy.
Statement 1 : \((p \rightarrow q) \leftrightarrow q \vee \sim p\) is a tautology.
Statement 2 : \(\sim(\sim p \wedge q) \wedge(p \vee q) \leftrightarrow p\) is a fallacy.
Show Hint
- Both statement 1 and statement 2 are true.
- Both statement 1 and statement 2 are false.
- Statement 1 is true and statement 2 is false.
- Statement 1 is false and statement 2 is true.
The Correct Option is C
Solution and Explanation
Statement 2: \(\sim(\sim p\wedge q)\equiv p\vee\sim q\). Then \((p\vee\sim q)\wedge(p\vee q)\equiv p\vee(\sim q\wedge q)\equiv p\vee F\equiv p\).
So the biconditional \(p\leftrightarrow p\) is always true, i.e. a tautology, not a fallacy. Statement 2 is false.
Answer: (3) Statement 1 is true and statement 2 is false.
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