Finding the Complement of a Boolean Expression The expression I was given was 
$$F=(a+c)(a+b')(a+b+c')$$
My first thought was to use DeMorgans so I expanded the two left parenthesis to the following:
$$(a+ac+b'c)(a+b+c')$$
Afterwards I applied DeMorgans to get the following:
$$ (\overline{a+ac+b'c}) + (\overline{a+b+c'})$$
which I brought to:
$$ 1+a'c'+bc'+a' +b' +c$$
I did the above using the following, $a+a' = 1$ and by applying the $\overline{x}$ to everything. This is the first complement I have done and I am not sure if this the correct way to do it? I was also wondering if there is a way to clean up the final answer that I got to. Than you very much.
 A: The negation of statements with pluses is a multiplication.
I.e.  $$\neg(a+b+c) = \neg(a)\neg(b)\neg(c)$$
So, your expression should be:
$$(\overline{a+ac+b'c + ab'}) + (\overline{a+b+c'})$$
$$= a'(a'+c')(b+c')(a'+b)+a'b'c$$
$$= (a' + a'c')(a'b+a'c'+b+bc')+a'b'c$$
$$= a'b+a'c+a'b+a'bc'+a'bc'+a'c'+a'bc'+a'bc'+a'b'c$$
$$= a'b+a'c+a'bc'+a'b'c$$
$$= a'(b(1+c') + c(1+b'))$$
$$= a'(b+c)$$
$$= a'b + a'c $$
A: One can easily simplify $F$ to $a$.  Hence the complement is $a'$.  To see that, first apply consensus to $(a+b')$ and $(a+b+c')$.  The result is $(a+c')$, which merges with $(a+c)$ to give $a$, which subsumes all other terms.
If you are not familiar with the consensus theorem, you can apply distributivity. Just observe that $a+ac=a$, so that $a+ac+b'c$ simplifies to $a+b'c$.  Also $b'cb$ and $b'cc'$ simplify to $0$.  Keep simplifying, and you'll get $a$ in the end.
You can also start with the application of De Morgan's Laws:
$$ F' = a'c' + a'b + a'b'c \enspace, $$
and then simplify.  With dual process to what you'd apply to $F$, you get $a'$ as result.
A: $$(a+c)(a+b')(a+b+c')\overset{Distribution}{=}$$
$$a+cb'(b+c') \overset{Reduction}{=}$$
$$a+cb'c' \overset{Complement}{=}$$
$$a+0 = a$$
