How to solve this boolean algebra problem?

Question

Given two expressions: $$A\bar{D}+A\bar{C}D +A\bar{B}C + ABCD = Y$$ and $$BD+A\bar{C}D=Z$$ is there a way to simplify this using the rules for Boolean Algebra? I tried different combinations, but if I factor out AC or D such as $$D(\bar{A}+A\bar{C})+AC(\bar{B}+BD)$$I am left with a problem that I cannot simplify further since I get multiple terms, the same goes for the second equation e.g. $$D(B+A\bar{C})$$ but then I am stuck. Any ideas?

Answer 1

The first equation is equivalent to $A = Y$: $$\begin{align} A\bar{D}+A\bar{C}D + A\bar{B}C + ABCD &= A(\bar{D}+\bar{C}D + C(\bar{B} + BD))\\ &= A(\bar{D} + \bar{C}D + C(\bar{B}(\bar{D}+D) + BD))\\ &= A(\bar{D} + \bar{C}D + C(\bar{B}\bar{D} + (\bar{B} + B)D))\\ &= A(\bar{D} + \bar{C}D + C\bar{B}\bar{D} + CD)\\ &= A(\bar{D} + (\bar{C}+C)D + C\bar{B}\bar{D})\\ &= A(\bar{D} + D + C\bar{B}\bar{D})\\ &= A(\top + C\bar{B}\bar{D})\\ &= A\top\\ &= A \end{align}$$