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I need to prove:

XY + ~XZ + YZ = XY + ~XZ

I cannot think how to do this.

I have tried factorising, but I just don't know of any rule that removes one of the terms like above.

I start with the LHS, naturally:

All I can do is get to:

Y(X+Z) + ~XZ

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    $\begingroup$ Since there are only three variables, writing out the two truth tables will do just fine. $\endgroup$ – David Richerby Jan 6 '15 at 23:19
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Here is an ugly solution, found with the Karnaugh table and the equation : A = A.B + A./B

XY + /XZ + YZ

= X.Y.Z + X.Y./Z + /X.Y.Z + /X./Y.Z + X.Y.Z + /X.Y.Z

(removing two redundant terms)

= X.Y.Z + X.Y./Z + /X./Y.Z + /X.Y.Z

= XY + /X.Z

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  • $\begingroup$ I'm afraid this won't be a valid solution, I've been staring at this blankly for some time now. The question simply asks to use 'Boolean algebra' to prove ... Anyother ideas? $\endgroup$ – lmsavk Jan 6 '15 at 20:01
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OK, answer is as follows, unsure if there's an easier method:

$XY+X'Z+YZ \\ XY+X'Z+YZ(1) \\ XY+X'Z+YZ(X+X') \\ XY+X'Z+XYZ+X'YZ \\ (XY+XYZ)+(X'Z+X'YZ) \\ XY(1+Z)+X'Z(1+Y) \\ XY(1)+X'Z(1) \\ XY+X'Z$

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Here is a simple argument:

To prove $XY+X'Z+YZ = XY+X'Z$, we only need to show that $$(YZ = 1) \Rightarrow (XY + X'Z = 1)$$ which is obvious, because $$(YZ = 1) \Rightarrow (Y = 1 \land Z = 1) \Rightarrow (XY + X'Z = 1)$$ no matter whether $X = 0$ or $X = 1$.

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Note that $XY+X'Z = Y + Z$.

Intuitively, because if X is true, then the second term must be false, which means whether the entire expression is true depends on $Y$. Likewise for $Z$. Then we get:

$Y + Z + YZ$

This statement is true when $X$ is true, or when $Y$ is true, or when both are true. But that is redundant, and we can reduce to:

$Y + Z$

Recall from earlier that we established that $XY+X'Z = Y + Z$. Thus we get:

$XY + X'Z$

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