If n is not an odd integer then square of n is not odd. Let P(n) be the predicate that is not an odd integer, and (n) be the predicate that the square of n is not odd. For direct proof we should prove ○‡n : (P(n) ⇒ Q(n)) ○Vn : (P(n) ⇒ Q(n)) ○Vn : (¬P(n) ⇒ ¬Q(n)) ○³n : (¬P(n) ⇒ ¬Q(n)) Ovn: (¬Q(n) ⇒ ¬P(n))

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