By Crama Y., Hammer P.

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**Example text**

Let f be a Boolean function and C1 , C2 be implicants of f . We say that C1 absorbs C2 if C1 ∨ C2 = C1 or, equivalently, if C2 ≤ C1 . 18. Let f be a Boolean function and C1 be an implicant of f . We say that C1 is a prime implicant of f if C1 is not absorbed by any other implicant of f (namely, if C2 is an implicant of f and C1 ≤ C2 , then C1 = C2 ). 17. 15. It is easy to verify that xy and xz are prime implicants of f , whereas xyz is not prime (since xyz ≤ xz). As a matter of fact, f has no prime implicants other than xy and xz.

The function represented by the CNF φ(x1 , x2 , . . , x2n ) = (x1 ∨ x2 )(x3 ∨ x4 ) . . 23 hereunder). , {x2n−1 , x2n }. Thus, ψ has 2n terms. Writing down all these terms requires exponentially large time and space in terms of the length of the original formula φ. 11 essentially shows that there is no hope of transforming an arbitrary expression (or even a CNF) into an equivalent DNF in polynomial time. In Chapter 4, we shall return to a finer discussion of the following, rather natural, but surprisingly difficult question: Given a DNF expression ψ of the function f , what is the complexity of generating a CNF expression of f ?

Let f be a Boolean function and C1 be an implicant of f . We say that C1 is a prime implicant of f if C1 is not absorbed by any other implicant of f (namely, if C2 is an implicant of f and C1 ≤ C2 , then C1 = C2 ). 17. 15. It is easy to verify that xy and xz are prime implicants of f , whereas xyz is not prime (since xyz ≤ xz). As a matter of fact, f has no prime implicants other than xy and xz. 7 Implicants and prime implicants 27 Prime implicants play a crucial role in constructing DNF expressions of Boolean functions.