Digital Electronics: Principles, Devices and Applications

192 Digital ElectronicsExample 6.2Simplify AB + CDA + BC + DSolution• Let AB + CD = X.• Then the given expression reduces to XX.• Therefore, AB + CDA + BC + D = 0.6.2 Postulates of Boolean AlgebraThe following are the important postulates of Boolean algebra:1. 11 = 1 0 + 0 = 0.2. 10 = 01 = 0 0 + 1 = 1 + 0 = 1.3. 00 = 0 1 + 1 = 1.4. 1 = 0 and 0 = 1.Many theorems of Boolean algebra are based on these postulates, which can be used to simplifyBoolean expressions. These theorems are discussed in the next section.6.3 Theorems of Boolean AlgebraThe theorems of Boolean algebra can be used to simplify many a complex Boolean expression andalso to transform the given expression into a more useful and meaningful equivalent expression. Thetheorems are presented as pairs, with the two theorems in a given pair being the dual of each other.These theorems can be very easily verified by the method of ‘perfect induction’. According to thismethod, the validity of the expression is tested for all possible combinations of values of the variablesinvolved. Also, since the validity of the theorem is based on its being true for all possible combinationsof values of variables, there is no reason why a variable cannot be replaced with its complement, orvice versa, without disturbing the validity. Another important point is that, if a given expression isvalid, its dual will also be valid. Therefore, in all the discussion to follow in this section, only one ofthe theorems in a given pair will be illustrated with a proof. Proof of the other being its dual is implied.6.3.1 Theorem 1 (Operations with ‘0’ and ‘1’)(a) 0X = 0 and (b) 1 + X = 1 (6.11)where X is not necessarily a single variable – it could be a term or even a large expression.Theorem 1(a) can be proved by substituting all possible values of X, that is, 0 and 1, into the givenexpression and checking whether the LHS equals the RHS:• For X = 0, LHS = 0.X = 0.0 = 0 = RHS.• For X = 1, LHS = 0.1 = 0 = RHS.Thus, 0.X = 0 irrespective of the value of X, and hence the proof.Theorem 1(b) can be proved in a similar manner. In general, according to theorem 1, 0.(Booleanexpression) = 0 and 1 + (Boolean expression) = 1. For example, 0AB + BC + CD = 0 and 1 +AB + BC + CD = 1, where A, B and C are Boolean variables.