I am honestly confused over here. I understand how bit manipulation works. But when it comes to Boolean Algebra I seem to be a bit stumped.
So if A = 1, then ~A = 0.
But in this example (~A * ~B) + (~C * ~D) = 0 + 0 or 0 | 0 = 0. But in this example it equals 1. Karnaugh Map
If $A=1$, then $\sim\! A=0$, sure. But also, if $A=0$, then $\sim\! A=1$. The $1$ in the K-map says that when all four inputs are false, you want the e segment of the display to lit up.
By the way, that box of the K-map corresponds to the minterm
$$ \sim\! A \,∗ \sim\! B\, ∗ \sim\! C\, ∗ \sim\! D \enspace,$$
which is true when all four inputs are false.