Can someone help me with simplifying this Boolean algebra
Prove: (x * y') + (y' * z) + (x' * z) = (x * y') + (x' * z)
x'= not x
Can you please show me step by step and the laws which you have applied in proving?
Thank you very much!
Can someone help me with simplifying this Boolean algebra
Prove: (x * y') + (y' * z) + (x' * z) = (x * y') + (x' * z)
x'= not x
Can you please show me step by step and the laws which you have applied in proving?
Thank you very much!
On
The Consensus Theorem does this in 1 step:
Consensus Theorem
$XY'+Y'Z+X'Z \Leftrightarrow XY'+X'Z$
We can derive the Consensus Theorem (and thus your equivalence as well) from some more basic principles:
$XY'+Y'Z+X'Z= \text{ Adjacency}$ (Adjacency says $PQ+PQ'=P$)
$XY'+XY'Z+X'Y'Z+X'Z= \text{ Absorption}$ ($XY'$ absorbs $XY'Z$ and $X'Z$ absorbs $X'Y'Z$)
$XY+X'Z$
If you didn't already know of Absorption and Adjacency, then you should immediately add them to your Boolean algebra toolbox, because they are super useful!
It's worth your while to study this proof because this pattern shows up a lot. In fact, it's the proof of an identity known as Consensus:
\begin{array}{l} & xy' + y'z + x'z & \text{ Given }\\ & xy' + y'z1 + x'z & \text{ Identity }\\ & xy' + y'z(x + x') + x'z & \text{ Identity }\\ & xy' + y'zx + y'zx' + x'z & \text{ Distributive }\\ & xy' + y'zx + x'z + y'zx' & \text{ Associative }\\ & xy'(1 + z) + x'z(1+ y') & \text{ Distributive }\\ & xy'1 + x'z1 & \text{ Identity }\\ & xy' + x'z & \text{ Identity }\\ \end{array}