(p ∧∼q)∧(∼p ∧ q) is ___.
A tautology
A fallacy
Both a tautology and a fallacy
Neither a tautology nor a fallacy
The AND connective is associative and commutative. Hence, (p ∧∼q)∧(∼p ∧ q)=(p ∧∼p)∧(∼q∧q) This statement is always false.
The statement (p⇒∼ p)∧(∼ p⇒p) is a: