Algebraic form from truth table with two outputs; simplifying boolean expression? I have a truth table like below:
x y z | a b
-----------
0 0 0 | 1 1
0 0 1 | 0 1
0 1 0 | 0 0
0 1 1 | 0 0
1 0 0 | 0 1
1 0 1 | 0 0
1 1 0 | 0 0
1 1 1 | 0 0

If the truth table had only one output, I would sum up the expressions that resulted in a true output, but I'm not entirely sure what to do for two outputs. Taking output $b$ for example:
$$b = \overline{xyz} + \overline{xy}z + \overline{x}yz$$
The complication arises with $a$ is introduced alongside $b$. Do I count only when both $a$ and $b$ are true?
$$ab = \overline{xyz}$$
Additionally, I'm not sure how to simplify an expression like this. Taking the expression for $b$ from earlier, is this how you do it?
$$\begin{align}b &= \overline{xyz} + \overline{xy}z + \overline{x}yz\\
&= \bar{x} + \bar{y} + \bar{z} + \bar{x} + \bar{y} + z + \bar{x} + y + z \text{ DeMorgan's Law}\\
&= \bar{x} + \bar{x} + \bar{x} + \bar{y} + \bar{y} + y + \bar{z} + z + z \text{ Commutation}\\
&= 3\bar{x} + \bar{y} + z\end{align}$$
The further I try to simplify, the less it makes sense in the context of circuitry.
 A: $$\begin{array}{ccc|cc} x & y & z & a & b \\ \hline
0 & 0 & 0 & 1 & 1 \\
0 & 0 & 1 & 0 & 1 \\
0 & 1 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 & 1 \\
1 & 0 & 1 & 0 & 0 \\
1 & 1 & 0 & 0 & 0 \\
1 & 1 & 1 & 0 & 0\end{array}$$ $$\implies \begin{cases}a= \overline x \cdot\overline y\cdot \overline z\\ b= \overline x \cdot\overline y\cdot \overline z+\overline x \cdot \overline y \cdot z + x\cdot\overline y \cdot \overline z \\ a+b = \overline x \cdot\overline y\cdot \overline z+\overline x \cdot \overline y \cdot z + x\cdot\overline y \cdot \overline z = b \\ a\cdot b = \overline x \cdot\overline y\cdot \overline z=a\end{cases}$$
Let's try to simplify $b$ a little:
$$\begin{align}b &= \overline x \cdot\overline y\cdot \overline z+\overline x \cdot \overline y \cdot z + x\cdot\overline y \cdot \overline z \\ &= \overline y \cdot (\overline x\cdot \overline z+\overline x \cdot z + x \cdot \overline z) \\ &= \overline y \cdot [\overline x\cdot (\overline z + z) + x \cdot \overline z] \\ &= \overline y \cdot (\overline x + x\cdot \overline z) \\ &= \overline y \cdot (\overline x\cdot (\overline z+1) + x\cdot \overline z) \\ &= \overline y \cdot (\overline x\cdot \overline z+\overline x + x\cdot \overline z) \\ &= \overline y \cdot ((\overline x +x)\cdot \overline z+\overline x) \\ &= \overline y \cdot (\overline z+\overline x) \\ &= \overline y \cdot \overline {z\cdot x}\end{align}$$
There's probably a slicker way of reaching that result, but it's been a while since I took electronics.  But this way works! ;)
