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Prove $x'z+xyz+xy'z=z$ can you show how you solve this using Boolean Algebra.

My main problem is when I do this $xz (y + y') = 1 $ So $1$ times $x =$ ?

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  • $\begingroup$ $$xz(y+y')=xz$$ and $x'z+xz=z(x'+x)=z$ $\endgroup$ Nov 10 '13 at 5:33
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$$ x'z + xyz + xy'z = z $$ $$ x'z + xz (y + y') \ // AssociationRule$$ $$ x'z + xz * 1 \ // \ a + a' = 1; a * 1 = a $$ $$ z (x'+x) \ // \ a + a' = 1$$ $$ z * 1 \ // \ a * 1 = a $$ $$ z $$

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How about simply $$ \begin{eqnarray} x'z + xyz + xy'z &=& (x'+xy+xy')z \\ &=&(x'+x(y+y'))z \\ &=&(x'+x)z \\ &=& z, \end{eqnarray} $$ using the distributive laws and $x+x'=1$.

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