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How can we prove $(x+yz)(y'+x)(y'+z')=x(y'+z')$ in a Boolean algebra $B$?

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2 Answers 2

The first two factors simplified to $$(x+yz)(y'+x)$$

$$=(x+yz)(x+y')$$ $$=x+\{(yz).y'\}$$ $$=x+0$$ $$=x$$

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How did you get from the second step to the third step? –  Pacerier May 7 '13 at 9:35

One way is proof by exhaustion:

\begin{matrix} x & y & z & (x+yz)(y'+x)(y'+z') & x(y'+z')\\ \hline False & False & False & False & False \\ False & False & True & False & False \\ False & True & False & False & False \\ False & True & True & False & False \\ True & False & False & True & True \\ True & False & True & True & True \\ True & True & False & True & True \\ True & True & True & False & False \end{matrix} Since the values match for any assignment of $(x,y,z)$, they are equal.

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Great illustration! + –  B. S. Jan 7 '13 at 9:22

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