 By Gallier J.

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Extra info for Discrete Mathematics for Computer Science Some Notes

Example text

BASICS CONCEPTS OF SET THEORY 51 Then, the subset axiom asserts the existence of a set Y so that for every x, x∈Y iff x ∈ A and P (X, x) which is equivalent to x∈Y Therefore, the set Y is our desired set, iff P (X, x). X. Observe that {A, B} = A ∩ B, {A1 , . . , An } = A1 ∩ · · · ∩ An . Note that ∅ is not defined. Intuitively, it would have to be the set of all sets, but such a set does not exist, as we now show. This is basically a version of Russell’s paradox. , there is no set to which every other set belongs.

This contradicts the fact that p2 +bpq +cq 2 = 0 and thus, finishes the proof. As as example of the proof by contrapositive method, we prove that if an integer n2 is even, then n must be even. Observe that if an integer is not even then it is odd (and vice-versa). Thus, the contrapositive of our statement is: If n is odd, then n2 is odd. But, to say that n is odd is to say that n = 2k + 1 and then, n2 = (2k + 1)2 = 4k 2 + 4k + 1 = 2(2k 2 + 2k) + 1, which shows that n2 is odd. 6. ADDING QUANTIFIERS; THE PROOF SYSTEMS NC⇒,∧,∨,∀,∃,⊥ , N G C 31 A real number a ∈ R is said to be irrational if it cannot be expressed as a number in Q √ (a fraction).

A typical example of redundancy is an elimination immediately follows an introduction, as in the following example in which D1 denotes a deduction with conclusion Γ, x : A → B and D2 denotes a deduction with conclusion Γ → A. D1 Γ, x : A → B Γ→A⇒B Γ→B D2 Γ→A Intuitively, it should be possible to construct a deduction for Γ → B from the two deductions D1 and D2 without using at all the hypothesis x : A. This is indeed the case. If we look closely at the deduction D1 , from the shape of the inference rules, assumptions are never created, and the leaves must be labeled with expressions of the form Γ , ∆, x : A, y : C → C or Γ, ∆, x : A → A, where y = x and either Γ = Γ or Γ = Γ , y : C.