Boolean Algebra Simplification Question - Proof of equation

Question

Boolean Algebra Simplification Question - Proof of equation
I`m trying to proof this equation:

X'.Y' + Y'.Z + X.Z + X.Y + Y.Z' = X'.Y'+X.Z+Y.Z'

What your are suggesting? to add some

(Y'+Y) or (Z'+Z)

Thanks.

EDIT - SOLVED

Answer 1

Start by expanding $X\cdot Y$:

$$X\cdot Y=X\cdot Y\cdot(Z+Z')=X\cdot Y\cdot Z+X\cdot Y\cdot Z'\;.$$

Now reduce $X\cdot Y+X\cdot Z+Y\cdot Z'$:

$$\begin{align*} X\cdot Y+X\cdot Z+Y\cdot Z'&=X\cdot Y\cdot Z+X\cdot Y\cdot Z'+X\cdot Z+Y\cdot Z'\\ &=(X\cdot Z+X\cdot Z\cdot Y)+(Y\cdot Z'+Y\cdot Z'\cdot X)\\ &=X\cdot Z+Y\cdot Z'\;. \end{align*}$$

That gets rid of the unwanted $X\cdot Y$ term and leaves only the $Y'\cdot Z$ term to be taken care of. You can get rid of it using the same idea.

Answer 2

Notice that $LHS = RHS \lor (\lnot Y \land Z) \lor (X \land Y)$. So the only way you might have an inequality is if $RHS$ is false and $LHS$ is true.

$LHS$ is false $\implies$ one of two options: Either $X=\text{True}, Y=\text{False}, Z=\text{False}$ or $X=\text{False}, Y=\text{True}, Z=\text{True}$.

In both cases, $X \land Y = \text{False}$ and $\lnot Y \land Z = \text{False}$, so $LHS$ remains false, and we have $LHS = RHS$.