For a better view, i'll skip *
for conjunction, and use '
for negation.
First you shall expand the 2 term disjunctions: Expand B*C
, A'*B'
and A'*C'
1) (A + A')BC + A'B'(C + C') + A'(B + B')C'
now distribute the parentheses.
2) ABC + A'BC + A'B'C + A'B'C' + A'BC' + A'B'C'
the fourth term and the last term are the same, A'B'C'
, so ignore one of them since p + p = p
or you can expand the situation for your needs (might be needed for some situations) as in p+p+p+p+....+p = p
3) So now, lets try to search for common terms. See the 2nd term and 5th term, A'BC
and A'BC'
. Take common parenthesis, A'B(C+C') => A'B
.
Do the same for 3rd term and the 4th term, A'B'C
and A'B'C'
. A'B'(C+C') => A'B'
since X+X' = 1
.
now we have:
ABC + A'B + A'B'
4) take common parenthesis again, 2nd and 3rd term: A'(B+B')
There you have ABC + A'
BC + A'B' + A'C' => ABC + A'