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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

Solution

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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.

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  • $\begingroup$ thanks. uploaded my solution. $\endgroup$ – Ofir Attia Apr 20 '13 at 12:06
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    $\begingroup$ @Ofir: You’re welcome, but I think that your last calculation is flawed: there’s no reason to conclude that $Y'ZZ=0$. Try instead $Y'Z=Y'Z(X+X')=XY'Z+X'Y'Z$, so that $Y'Z+X'Y'+XZ=XY'Z+X'Y'Z+X'Y'+XZ=X'Y'+XZ$ by the same kind of absorption that I used to eliminate $XY$. $\endgroup$ – Brian M. Scott Apr 20 '13 at 12:21
  • $\begingroup$ yes, i see now i tought it ZZ' . thanks. $\endgroup$ – Ofir Attia Apr 20 '13 at 12:33
  • $\begingroup$ @Ofir: You’re welcome; I’m glad we got it straightened out. $\endgroup$ – Brian M. Scott Apr 20 '13 at 12:34
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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 $LHS = RHS$.

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