A Boolean Algebra simplification problem

Question

XY + (YZX)' + YZ

So above I have this boolean expression I have done the work out and I got 1. I was wondering if that was the correct answer or did I miss a step.

Yes, it is 1, because (xyz)' is 1 for any choice of (x,y,z) besides (1,1,1).... and (1,1,1) makes true the part xy (or yz, if you prefer). — user376343, Sep 22 '19 at 14:30

score 0 · Answer 1 · answered Sep 22 '19 at 15:23

Let $X,Y,Z$ be subsets of $A$.

Then checking whether the expression equals $1$ comes to the same as checking that:$$(X\cap Y)\cup(Y\cap Z\cap X)^{\complement}\cup(Y\cap Z)=A$$or equivalently:$$A\subseteq(X\cap Y)\cup(Y\cap Z\cap X)^{\complement}\cup(Y\cap Z)$$It is not really difficult to verify this statement.

score 0 · Answer 2 · answered Sep 23 '19 at 02:40

0

Using

Reduction

$P'+PQ=P'+Q$

We get:

$XY+(YZX)'+YZ= XY + Y' + Z' + X' +YZ=X+Y'+Z'+X'+Z=(X+X')+(Z+Z')+Y=1+1+Y=1$

answered Sep 23 '19 at 02:40

Bram28

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A Boolean Algebra simplification problem

2 Answers2