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The statement ~(p ⇒ q) ⇔ (~p ∨ ~q) is:

1. A tautology
2. A contradiction
3. Neither a tautology nor a contradiction
4. No definite conclusion can be drawn

Correct answer: Neither a tautology nor a contradiction

Solution

~(p=>q) is equivalent to p AND ~q. Compare with (~p OR ~q): for p=T,q=F both sides are true; for p=F,q=F left is false while right is true. Since the biconditional is true for some rows and false for others, the statement is neither a tautology nor a contradiction.

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