F(A, B, C, D) = A'B'C' + AC' + ACD + ACD' + A'B'D'
I first did AC(D+D') which got me just AC, and then A(C+C') which left me with just A. The function is now A'B'C'+ A + A'B'D'. I am unsure if I can do things like pulling the A out of A'B'D' to do A+A. If it isn't, I am stuck, what more can I do?
$$A+A'B'D' \overset{Distribution}{=}$$
$$(A+A')(A+B')(A+D') \overset{Complement}{=}$$
$$1(A+B')(A+D') \overset{Identity}{=}$$
$$(A+B')(A+D') \overset{Distribution}{=}$$
$$A+B'D'$$