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# On the Complexity of Slice Functions,” Theoretic Computer Science.

Methods of Proof In this section we briefly introduce several methods of proof that are used in this book, namely, proof by induction, proof by contradiction, and the pigeonhole principle. In the previous c John E Savage 1.3 Methods of Proof 15 section we saw proof by reduction: in each step the condition to be established was translated into another condition until a condition was found that was shown to be true. Proofs by induction use predicates, that is, functions of the kind P : → B. The truth value of the predicate P : → B on the natural number n, denoted P(n), is 1 or 0 depending on whether or not the predicate is True or False. Proofs by induction are used to prove statements of the kind, “For all natural numbers n, predicate (or property) P is true.” Consider the function S1 : → defined by the following sum: S1(n) = n j=1 j (1.1) We use induction to prove that S1(n) = n(n + 1)/2 is true for each n ∈ . DEFINITION 1.3.1 A proof by induction has a predicate P, a basis step, an induction hypothesis, and an inductive step. The basis establishes that P(k) is true for integer k. The induction hypothesis assumes that for some fixed but arbitrary natural number n ≥ k, the statements P(k), P(k + 1), … , P(n) are true. The inductive step is a proof that P(n + 1) is true given the induction hypothesis. It follows from this definition that a proof by induction with the predicate P establishes that P is true for all natural numbers larger than or equal to k because the inductive step establishes the truth of P(n + 1) for arbitrary integer n greater than or equal to k. Also, induction may be used to show that a predicate holds for a subset of the natural numbers. For example, the hypothesis that every even natural number is divisible by 2 is one that would be defined only on the even numbers. The following proof by induction shows that S1(n) = n(n + 1)/2 for n ≥ 0. LEMMA 1.3.1 For all n ≥ 0, S1(n) = n(n + 1)/2. Proof PREDICATE: The value of the predicate P on n, P(n), is True if S1(n) = n(n + 1)/2 and False otherwise. BASIS STEP: Clearly, S1(0) = 0 from both the sum and the closed form given above. INDUCTION HYPOTHESIS: S1(k) = k(k + 1)/2 for k = 0, 1, 2, … , n. INDUCTIVE STEP: By the definition of the sum for S1 given in (1.1), S1(n+1) = S1(n)+ n + 1. Thus, it follows that S1(n + 1) = n(n + 1)/2 + n + 1. Factoring out n + 1 and rewriting the expression, we have that S1(n + 1)=(n + 1)((n + 1) + 1)/2, exactly the desired form. Thus, the statement of the theorem follows for all values of n. We now define proof by contradiction. DEFINITION 1.3.2 A proof by contradicti