CMPUT272 Lecture Notes - Lecture 11: Prime Number, Boolean Satisfiability Problem, Propositional Calculus

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Let p(x) be defined in k p(k+1) - inductive step then a, p(n) Thereom: for all integers n -> 4, and an integer. Now (2^(k+1)) = (2 * 2^k) < (2 * (k+1)!) Proves up to some "b" - modus pones. And 2. i, a p(k) (p(a) p(b+1) (p(a) p(b + 1)) n n, sn s0 s1 s2 k>b( i < k, Now sk = (s(k-1) + s(k-2) + s(k-3)) Now sk = (s(k-1) + s(k-2) + s(k-3)) (3^k-1 + 3^k-1 + 3^k-1) = 3 * 3^k-1 = 3^k (3^k-1 + 3^k-1 + 3^k-1) = 3 * 3^k-1 = 3^k. Every nonempty set of non-negative integers has a. Every nonempty set of non-negative integers has a smallest element. smallest element. By being unable to cronstruct a counter-example in the. 3sat: (x y z) x ( 2 is a product of. 3sat: (x y z) x ( 2 is a product of primes primes.

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