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Identify the statement that is incorrect.

1. The implication p → q is equivalent to the disjunction ~p ∨ q.
2. When p, q, r take the values T, F, T respectively, the expression (p ∨ q) ∧ (q ∨ r) evaluates to T.
3. ~(p ∨ q ∨ r) is logically equivalent to ~p ∧ ~q ∧ ~r.
4. The proposition p ∧ ~(p ∨ q) is always T.

Correct answer: The proposition p ∧ ~(p ∨ q) is always T.

Solution

Take p=T: ~(p∨q)=F, so p∧F=F; with p=F the whole thing is F too. Hence p∧~(p∨q) is always FALSE, so the claim that it is always T is the incorrect statement. (A is De Morgan-valid, B evaluates to T, C is De Morgan-valid.)

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