I was working through a problem for a Computer Engineering course and i was given this logic function
F(A,B,C,D) = ~A~BC~D + ~AB~C~D + ~ABC~D + ABC~D
After factoring out BC~D from two of the terms and getting (~A + A) = 1 *(BC~D) I got
F(A,B,C,D) = ~A(~BC~D + B~C~D) + BC~D
Is there any simplification that can be done any farther?
Yes, it can be simplified a little further. For convenience I’ll write $A'$ instead of $\sim\!\!A$.
$$\begin{align*} A'B'CD'&+A'BC'D'+\color{red}{A'BCD'}+ABCD'\\ &=A'B'CD'+A'BC'D'+\color{red}{A'BCD'+A'BCD'}+ABCD'\\ &=(A'B'CD'+\color{red}{A'BCD'})+A'BC'D'+(\color{red}{A'BCD'}+ABCD')\\ &=A'(B'+B)CD'+A'BC'D'+(A'+A)BCD'\\ &=A'CD'+A'BC'D'+BCD'\\ &=(A'C+A'BC'+BC)D'\\ &=\big(A'(\color{red}C+BC')+BC\big)D'\\ &=\big(A'(\color{red}{C+BC}+BC')+BC\big)D'\\ &=\left(A'\big(C+B(C+C')\big)+BC\right)D'\\ &=\big(A'(C+B)+BC\big)D'\\ &=A'BD'+A'CD'+BCD' \end{align*}$$
