According to WolframAlpha, $$(A_3\cdot(A_2+A_1+A_0))+(\overline{A_3}\cdot A_2 \cdot\overline{A_1}\cdot\overline{A_0})$$ should simplify to $$=A_3\cdot A_0+A_3\cdot A_1+A_2\cdot\overline{A_1}\cdot\overline{A_0}$$.
But how? Here is my attempt:
Specifically, I don't know how to get from $$A_2\cdot(A_3+\overline{A_1}\cdot\overline{A_0})$$
to
$$A_2\cdot\overline{A_1}\cdot\overline{A_0}$$
Did I miss something? Any help is appreciated!
Best Answer
Obviously, the subexpression $A_2(A_3+A_1'A_0')$ is not equivalent to $A_2A_1'A_0'$
So, did you do something wrong?
No, you did everything right.
It is the fact that this subexpression appears in the context of the larger expression as a whole that you can simplify it.
Indeed, how in the step before it you yourself go from $A_3+A_3'A_1'A_0'$ to $A_3+A_1'A_0'$ ... Whoa, is that because $A_3'A_1'A_0'$ is equivalent to $A_1'A_0'$? No, of course not! Rather, it is the fact that you also have $A_3$ that you can reduce the expression $A_3'A_1'A_0'$ to $A_1'A_0'$
Something similar is going on here well: we'll need the fact that we also have $A_3A_1$ and $A_3A_0$ in order to simplify $A_2(A_3+A_1'A_0')$ to $A_2A_1'A_0'$
This is not easy though, as you can see from the Answer by @Manx
Still, using the Consensus Law, it can be made a little easier:
\begin{align} &A_2\cdot (A_3+(\overline{A_1}\cdot\overline{A_0}))+A_3\cdot A_0+A_3\cdot A_1\\ =&A_2\cdot A_3+A_2\cdot\overline{A_1}\cdot\overline{A_0}+A_3\cdot A_0+A_3\cdot A_1\tag*{Distributive law}\\ =&A_2\cdot A_3+A_2\cdot\overline{A_1}\cdot\overline{A_0}+A_3\cdot (A_0+A_1)\tag*{Distributive law}\\ =&A_2\cdot A_3+A_2\cdot\overline{A_1+A_0}+A_3\cdot (A_0+A_1)\tag*{DeMorgan}\\ =&A_2\cdot\overline{A_1+A_0}+A_3\cdot (A_0+A_1)\tag*{Consensus}\\ =&A_2\cdot\overline{A_1}\cdot\overline{A_0} + A_3\cdot A_0+A_3\cdot A_1\tag*{DeMorgan, Distribution}\\ \end{align}