Question

Among the two statements
(S1) : (p ⇒ q)∧(q ∧(~q)) is a contradiction and  
(S2) : (p∧q) v ((~p)∧q) v (p ∧ (~q)) v ((~p) ∧ (~q)) is a tautology 

• A. both are true
• B. both are false
• C. only (S1) is true
• D. only (S2) is true

Hint:

Remember the truth tables for basic logical operations (and, or, not, implication). A contradiction is always false, and a tautology is always true

Answer 1

The Correct Option is A

For \( S_1 : (p \rightarrow q) \land (p \land \sim q) \), the truth table is: 

\( p \)	\( q \)	\( p \rightarrow q \)	\( p \land \sim q \)	\( S_1 \)
T	T	T	F	F
T	F	F	T	F
F	T	T	F	F
F	F	T	F	F

From the truth table, \( S_1 \) is a Contradiction, as \( S_1 \) is always false.

For \( S_2 : (\sim p \land q) \lor (p \land \sim q) \), the truth table is:

\( p \)	\( q \)	\( p \land q \)	\( \sim p \land q \)	\( p \land \sim q \)	\( (\sim p \land q) \lor (p \land \sim q) \)	\( S_2 \)
T	T	T	F	F	F	F
T	F	F	F	T	T	T
F	T	F	T	F	T	T
F	F	F	F	F	F	F

From the truth table, \( S_2 \) is a Tautology, as \( S_2 \) is always true.

Hence:

\( S_1 \) is Contradiction, and \( S_2 \) is Tautology.