Question:
(S1): (p⇒q)∧(p∧((~q)) is a contradiction and
(S2): (p∧q)v((~q)v((~p)∧ q)v (p∧(~q))v((~p)∧ (~q)) is a tautology
(S1): (p⇒q)∧(p∧((~q)) is a contradiction and
(S2): (p∧q)v((~q)v((~p)∧ q)v (p∧(~q))v((~p)∧ (~q)) is a tautology
(S2): (p∧q)v((~q)v((~p)∧ q)v (p∧(~q))v((~p)∧ (~q)) is a tautology
Updated On: Nov 21, 2025
- both (S1) and (S2) are true.
- both (S1) and (S2) are false.
- only (S1) is true.
- only (S2) is true.
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The Correct Option is A
Solution and Explanation
The correct option is(A): both (S1) and (S2) are true
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