Simplify f by Boolean algebra as much as possible fa b c ab

Simplify f by Boolean algebra as much as possible.

   f(a, b, c) = abc + ab’c + a’bc +a’b’c + a’bc’

Solution

given f(a,b,c) = abc+ ab\'c+a\'bc +a\'b\'c +a\'bc\'

= ac(b+b\') +a\'bc+a\'b\'c+a\'bc\'

=ac(1) +a\'bc+a\'b\'c+a\'bc\' [ Formula: b+b\' =1]

=ac+a\'bc+a\'b\'c+a\'bc\'

= ac+a\'bc+a\'(b\'c+bc\')

=ac+a\'bc+a\'(b XOR c) [since it is an XOR operation ]