Question

Among the statements : 
\((S1)\) \((( p \vee q ) \Rightarrow r ) \Leftrightarrow( p \Rightarrow r )\)
\((S2)\)\((( p \vee q ) \Rightarrow r ) \Leftrightarrow(( p \Rightarrow r ) \vee( q \Rightarrow r ))\)

• A. only (S₂) is a tautology
• B. both (S₁) and (S₂) are tautologies
• C. only (S₁) is a tautology
• D. neither (S₁) nor (S₂) is a tautology

Answer 1

The Correct Option is D

To solve this problem, we need to evaluate whether the given logical statements \((S1)\) and \((S2)\) are tautologies. A tautology is a formula that is true in every possible interpretation.

Let's examine each statement individually:

1. Statement \((S1)\): \((( p \vee q ) \Rightarrow r ) \Leftrightarrow( p \Rightarrow r )\)
   • The expression \(( p \vee q ) \Rightarrow r\) can be rewritten as \((\neg ( p \vee q ) \vee r)\), which further simplifies to \(((\neg p \wedge \neg q) \vee r)\).
   • The expression \((p \Rightarrow r)\) can be rewritten as \((\neg p \vee r)\).
   • The statement is examining whether these two expressions are logically equivalent in all conditions. Upon testing with possible truth values for \(p\), \(q\), and \(r\), this will not hold true in all cases. Therefore, \((S1)\) is not a tautology.
2. Statement \((S2)\): \((( p \vee q ) \Rightarrow r ) \Leftrightarrow(( p \Rightarrow r ) \vee( q \Rightarrow r ))\)
   • We already have \((( p \vee q ) \Rightarrow r) = (\neg (p \vee q) \vee r)\) as derived above.
   • The expression \(( (p \Rightarrow r) \vee (q \Rightarrow r) )\) is equivalent to \(((\neg p \vee r) \vee (\neg q \vee r))\).
   • Check whether these are logically equivalent for all possible combinations of truth values. Not all truth combinations satisfy this equivalence, meaning \((S2)\) is also not a tautology.

Therefore, neither statement \((S1)\) nor statement \((S2)\) holds as a tautology. This validates the correct answer choice:

Correct Answer: neither \((S1)\) nor \((S2)\) is a tautology