# How can this expression be simplified using boolean algebra?

So I have this expression: $$A'BD + B'C' + AC' + C'D$$ How can I simplify it using boole's algebra? I figure out the $C'D$ is included in the other terms through Karnaugh maps but I can't figure out the simplification.

To show that $C'D \leq AC' + A'BD + B'C'$ is equivalent to show that their product is $C'D$.
Now, \begin{align} C'D ( AC' + A'BD + B'C' ) &= AC'D + A'BC'D + B'C'D \\ &= (A + A'B + B') C'D. \end{align} Now we prove that $A + A'B + B' = 1$, concluding the initial inequality. \begin{align} A + A'B + B' &= A + A'B + (A + A')B' \\ &= A + A'B + A'B' \\ &= A + A'( B + B') \\ &= A + A' \\ &= 1. \end{align}
• Just a request, if you did not know from karnaugh maps that $C'D$ was included how would you simplify this expression? I was thinking about expanding all terms and cut equal ones but there ought to be a more beautiful solution. Right? @amrsa – Carlos Sá Nov 3 '16 at 20:16