Download Advanced Formal Verification by Rolf Drechsler PDF

By Rolf Drechsler

Advanced Formal Verification indicates the newest advancements within the verification area from the views of the person and the developer. international top specialists describe the underlying tools of latest verification instruments and describe a variety of situations from business perform. within the first a part of the publication the center options of present day formal verification instruments, resembling SAT and BDDs are addressed. additionally, multipliers, that are recognized to be tough, are studied. the second one half supplies perception in expert instruments and the underlying method, comparable to estate checking and statement established verification. eventually, analog parts need to be thought of to deal with whole method on chip designs.

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Xn . Denote by B the set {0, 1} of values taken by a Boolean variable. ,xn . A point of B n is an assignment of values to all n variables. 19 Let p be a point of the Boolean space falsifying a clause C. The 1-neighborhood of the point p with respect to the clause C (written Nbhd(p,C)) is the set of points that are at Hamming distance 1 from p and that satisfy C. 11 It is not hard to see that the number of points in Nbhd(p, C) is equal to that of literals in C. 4 Let C = x1 ∨ x3 ∨ x6 be a clause specified in the Boolean space of 6 variables x1 , .

The set of coding variables v(B) = {q1 , q2 } consists of two Boolean variables. The Boolean vector 01 where q0 = 0,q1 = 1 is the code of b1 under the encoding q. In the following exposition we make the assumptions below. 1 Each gate of a Boolean circuit and each block of a specification has two inputs and one output. 2 The multiplicity of each primary input (or output) variable of a specification is a power of 2. 4 If a1 and a2 are values of a variable A of a specification and a1 = a2 , then q(a1 ) = q(a2 ).

Proof Assume the contrary. e. F (p∗ ) = 1. It is not hard to see that p∗ ∈ / P because each point p ∈ P is assigned a clause C = g(p) such that C(p)=0 and so F (p)=0. Let p be a point of P that is the closest to p∗ in Hamming distance. e. C = g(p). Denote by Y the set of variables values of which are different in p and p∗ . Let us show that C can not have literals of variables of Y . e. that C contains a literal of x ∈ Y . Then, since P is stable with respect to F and g, it has to contain the point p which is obtained from p by flipping the value of x.

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