Boolean Algebra

Consensus Theorem

Consensus theorem:

Given a pair of terms for which a variable appears in one term and its compliment in the other term then consensus term is formed by ANDing the original terms together leaving out the selected variable and its compliment.

e.g.        Find consensus term out of the two terms X.Y & X’.Z

                Consensus term is Y.Z

e.g.        Find consensus term out of the two terms  XYZ & Y’ZW’

Consensus term is (XZ). (ZW’)

SO X is left and simplified expression is XY’ + Y

And apply elimination law we get answer as X + Y

Or apply distributive law so we get (X + Y) (Y’ + Y)

And Y + Y’ = 1 hence we get simplified answer as X + Y

Leaving the consensus term from expression and we get the result as (X +Y) (X’+ Z)

Min term: All the combinations of the n variable to form 2 n AND terms are called MIN TERMs or standard products. For n=2 and variables x & y we have min terms as xy, xy’, x’y, x’y’ and similarly for n=3 and variables x, y, z we have the min terms as xyz, xyz’, x’yz, xy’z, x’y’z, x’yz’, xy’z’, x’y’z’. Min term is generally represented by letter ‘m’ (lower case) and to distinguish between different min terms we place a subscript with m and suffix is decided for a particular term by following procedure:

We put a 0 for the literal with compliment (‘) and a 1 for the literal without compliment and then take its binary equivalent. Decimal we get is placed as a subscript as shown below

xyz = 111 = 7 so min term is m7                   x’yz’ = 010 = 2    m2                       xyz’ = 110 = 6   metc

Max terms: similarly all the combinations of the n variable to form 2 n OR terms are called MAX TERMs or standard sums. For n=2 and variables x & y we have max terms as x+y, x+y’, x’+y, x’+y’ and similarly for n=3 and variables x, y, z we have the max terms as x+y+z, x+y+z’, x’+y+z, x+y’+z, x’+y’+z, x’+y+z’, x+y’+z’, x’+y’+z’. Max term is represented by ‘M’ (capital) and procedure to find the subscript is exact opposite.

We put a 1 for the literal with compliment (‘) and a 0 for the literal without compliment and then take its binary equivalent. Decimal we get is placed as a subscript as shown below

x+y’+z= 010 = 2 max term is M2                          x’+y+z =100 = 4 M4                                    x’+y’+z’=111=7  M7

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