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Among the statements (S1): (p ⇒ q) ∨ ((~p) ∧ q) is tautology (S2): (q ⇒ p) ∨ ((~p) ∧ q) is contradiction

1. Neither (S1) and (S2) is True
2. Both (S1) and (S2) are True
3. Only (S2) is True
4. Only (S1) is True

Correct answer: Neither (S1) and (S2) is True

Solution

Both statements are not true; (S1) is not a tautology as it can be false under certain conditions, and (S2) is not a contradiction since it can also be true depending on the truth values of p and q.

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