Question

simplify the following Boolean expression to a minimum number of literals(a+b+c')(a'b'+c)​

Answer 1

(a+b+c')(a'b'+c)

=aa' ab' ac + ba bb' bc + c'a' c'b' c'c

=1 ab' ac + ba 1 bc + c'a' c'b' 1

=1 + 1 + a' + b' + c' + ac + bc + 1

=a'b'c' + ac + bc

=a'b'c'+c(a+b)