# Boolean Algebra Simplifying complex equation

I am trying to simplify the following equation and I am getting stuck on a line and I can't cut it down any further. I'm not sure if certain 'moves' are legal or not.

F(A,B,C,D) = A'B'C' +ACD + A’BCD + A’BC’ + ABC + AB’CD'

After regrouping I got: A'(B'C' + BCD + BC') + A(CD + BC + B'CD')

I thought right after this you'd get 1 because A + A' is 1 and then after that you would get each term in parentheses leftover as the final expression. Is that right?

• The conversation about whether this question is off-topic has been moved to chat. – D.W. Oct 16 '18 at 17:21

Whenever you are faced with two Boolean expressions $$f,g$$ on $$n$$ variables and wish to know whether they are equivalent, there is a simple algorithm you can apply:
Go over all $$2^n$$ possible truth assignments, and check whether $$f$$ and $$g$$ have the same truth value on each.
While this is infeasible for large $$n$$, in your case $$n = 4$$, so there are only 16 many truth assignments to check; in fact, you could even do this by hand, though it might be less error-prone if you programmed it.