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By Crama Y., Hammer P.

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Extra info for Boolean functions. Theory, algorithms, and applications

Example text

Fm are m Boolean functions on B n , then the Boolean function ψ(f1 , f2 , . . , fm ) can be defined in the natural way: Namely, for all X ∗ ∈ Bn , ψ(f1 , f2 , . . , fm ) (X ∗ ) = ψ(f1 (X ∗ ), f2 (X ∗ ), . . 2) simply boils down to function composition). 3 Duality 13 Boolean functions on B n , then the functions f ∨ g, f ∧ g, and f are defined, for all X∗ ∈ Bn , by (f ∨ g)(X∗ ) = f (X ∗ ) ∨ g(X∗ ), (f ∧ g)(X∗ ) = f (X ∗ ) ∧ g(X∗ ), f (X ∗ ) = f (X ∗ ). 8. The dual of a Boolean function f is the function f d defined by f d (X) = f (X) for all X = (x1 , x2 , .

6. We say that two Boolean expressions ψ and φ are equivalent if they represent the same Boolean function. When this is the case, we write ψ = φ. 1 are equivalent even though they are not identical. 5. The function f (x, y, z) represented by ψ1 (x, y, z) = (x ∨ y)(y ∨ z) ∨ xyz (see previous examples) is also represented by the expression φ = x z ∨ y. Indeed, (x ∨ y)(y ∨ z) ∨ xyz = (xy ∨ x z ∨ yy ∨ yz) ∨ xyz = xy ∨ x z ∨ y ∨ yz ∨ xyz = xz∨y (distributivity) (idempotency and associativity) (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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