By Crama Y., Hammer P.

ISBN-10: 0521847516

ISBN-13: 9780521847513

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Example text

2. We can compute, for instance: fφ1 (0) = 0, fφ1 (1) = 1, fφ2 (0) = 0 = 1, fφ2 (1) = 1 = 0, fψ1 (0, 0, 0) = (((0 ∨ 0)(0 ∨ 0)) ∨ ((00)0)) = 1, . . 2 represents the function f , where f (0, 0, 1) = f (1, 0, 0) = f (1, 0, 1) = 0, f (0, 0, 0) = f (0, 1, 0) = f (0, 1, 1) = f (1, 1, 0) = f (1, 1, 1) = 1. Thus, we can write f (x, y, z) = ψ1 (x, y, z) = (((x ∨ y)(y ∨ z)) ∨ ((xy)z)). Remark. So that we can get rid of parentheses when writing Boolean expressions, we assume from now on a priority ranking of the elementary operations: Namely, we assume that disjunction has lower priority than conjunction, which has lower priority than complementation.

X n ). Let us enumerate some of the elementary properties of disjunction, conjunction, and complementation. 1. For all x, y, z ∈ B, the following identities hold: (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) x ∨ 1 = 1 and x ∧ 0 = 0; x ∨ 0 = x and x ∧ 1 = x; x ∨ y = y ∨ x and x y = y x (commutativity); (x ∨ y) ∨ z = x ∨ (y ∨ z) and x (y z) = (x y) z (associativity); x ∨ x = x and x x = x (idempotency); x ∨ (x y) = x and x (x ∨ y) = x (absorption); x ∨ (y z) = (x ∨ y) (x ∨ z) and x (y ∨ z) = (x y) ∨ (x z) (distributivity); x ∨ x = 1 and x x = 0; x = x (involution); (x ∨ y) = x y and (x y) = x ∨ y (De Morgan’s laws); x ∨ (x y) = x ∨ y and x (x ∨ y) = x y (Boolean absorption).

It is easy to check that φ is equivalent to the CNF (x ∨ y)(y ∨ z) with clauses (x ∨ y) and (y ∨ z). The expression ψ2 (x1 , x2 , x3 , x4 ) = x1 x2 ∨ x 3 x 4 is a DNF; it is equivalent to the CNF (x1 ∨ x 3 )(x1 ∨ x 4 )(x2 ∨ x 3 )(x2 ∨ x 4 ). 4) = (x ∨ y)(y ∨ z). 3) (which is not a normal form). This is not an accident. Indeed, we can now establish a fundamental property of Boolean functions. 4. Every Boolean function can be represented by a disjunctive normal form and by a conjunctive normal form.

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