I'm having a very difficult time proving the following 3 expressions:
$$\begin{align*}
&x\cdot y\cdot z+x'\cdot z=y\cdot z+x'\cdot z\\
&x\cdot y+y\cdot z+x'\cdot z=x\cdot y+x'\cdot z\\
&(x'\cdot z'+x'\cdot y+x'\cdot z+x\cdot y)'=x\cdot y'
\end{align*}$$
I need to show what law/theorem/postulate is used for each step of the proof – and I don't even know where to start. The last time I did any sort of algebra was at least 7 years ago, and even then it was very basic.
Being thrown into Boolean algebra, only provided a sheet with all the theorems/etc… is proving to be impossible for me. I understand the methodology behind mathematical proofs and boolean simplification, I just don't see what theorems/laws can be used when I look at these….
Any help would be greatly appreciated here!
Best Answer
Essentially, we manipulate the equation in whatever way possible, trying to make the left-hand side look like the right-hand side. There's no real magic method here, we just keep trying until we succeed.
In the second example, we can do:
$\small \begin{align*} x\cdot y+y\cdot z+x'\cdot z &= x\cdot y + 1 \cdot (y\cdot z)+x'\cdot z & \text{identity for } \cdot \\ &= x\cdot y + (x+x') \cdot (y\cdot z)+x'\cdot z & \text{complementation } x+x'=1\\ &=x\cdot y + x\cdot (y\cdot z)+x'\cdot (y\cdot z)+x'\cdot z & \text{distributivity} \\ &=x\cdot y + (x\cdot y)\cdot z+x'\cdot (y\cdot z)+x'\cdot z & \text{associativity} \\ &=(x\cdot y) \cdot 1 + (x\cdot y)\cdot z+x'\cdot (y\cdot z)+x'\cdot z & \text{identity for } \cdot \\ &=(x\cdot y) \cdot (1 + z)+x'\cdot (y\cdot z)+x'\cdot z & \text{associativity} \\ &=(x\cdot y) \cdot (1 + z)+x'\cdot (y\cdot z)+x'\cdot z & \text{distributivity} \\ &=(x\cdot y) \cdot 1+x'\cdot (y\cdot z)+x'\cdot z & \text{annihilator for } + \\ &=x\cdot y+x'\cdot (y\cdot z)+x'\cdot z & \text{identity for } \cdot \\ &=x\cdot y+(x'\cdot z) \cdot y+x'\cdot z & \text{assoc. and comm. of } \cdot \\ &=x\cdot y+(x'\cdot z) \cdot y+(x'\cdot z) \cdot 1 & \text{identity for } \cdot \\ &=x\cdot y+(x'\cdot z) \cdot (y+1) & \text{distributivity} \\ &=x\cdot y+x'\cdot z & \text{annihilator for } +, \\ \end{align*} $
as desired.