So I have been given the following expression, but I cannot seem to solve it, can anyone do this and show the steps please?
Prove XY'Z + XYZ' + XYZ = XY + XZ
XY'Z + XYZ' + XYZ = XY + XZ
- Notice X and Z are common factors between XY'Z and XYZ.
XZ(Y' + Y) + XYZ' =
- Y' + Y is equal to 1 (if Y=0 then Y'=1 and so 0 + 1 = 1, that is 0 or 1 = 1. Similarly, if Y=1 then Y'=0 and so 1 + 0 = 1). Therefore, what you get is:
XZ·1 + XYZ' =
- XZ·1 = XZ since A·1 = A (if A=0 then 0·1 is 0 and if A=1 then 1·1 = 1). Now the function is simplified to:
XZ + XYZ' =
- Notice once again X is a common factor between XZ and XYZ'.
X(Z + YZ') =
- Notice this time that Z + YZ' is a special case of the distributive law, which is A + A'B = A + B. This is because if we apply the general distributive law A + BC = (A + B)·(A + C) then we get A + A'B = (A + A')·(A + B) = 1·(A + B) = A + B. Following this reasoning we get to simplify the function even further:
X(Z + Y) =
- All that's left is for us to use the distributive law and we finally arrive to the final result:
XY + XZ
Please note that nothing is written between variables, an AND operator (or "·" symbol) is assumed. It's just a way to save space.