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

We will evaluate the truth values of the given statements \(S_1\) and \(S_2\) for different truth values of \(p\), \(q\), and \(r\). The truth table is as follows:

\(p\)	\(q\)	\(r\)	\(p \lor q\)	\((p \lor q) \Rightarrow r\)	\(p \Rightarrow r\)	\((p \lor q) \Rightarrow r \Leftrightarrow p \Rightarrow r\)	\((p \lor q) \Rightarrow r \Leftrightarrow (p \Rightarrow r) \lor (q \Rightarrow r)\)
T	T	T	T	T	T	T	T
T	T	F	T	F	F	T	T
T	F	T	T	T	T	T	T
T	F	F	T	F	F	T	T
F	T	T	T	T	T	T	T
F	T	F	T	F	T	T	T
F	F	T	F	T	T	T	T
F	F	F	F	T	T	T	T

Analysis of Statements

For statement \(S_1\), the truth table shows that \((p \lor q) \Rightarrow r \Leftrightarrow p \Rightarrow r\) is not true for all cases. Specifically, it is false when \(p = \text{T}, q = \text{T}, r = \text{F}\).

For statement \(S_2\), \((p \lor q) \Rightarrow r \Leftrightarrow (p \Rightarrow r) \lor (q \Rightarrow r)\) is also not a tautology because it is not true in all cases.

Conclusion

Neither \(S_1\) nor \(S_2\) is a tautology.

Therefore, the correct answer is:

\[ \boxed{4}. \]

Answer 2