How does the equation:
(x * z) + (!x * y) + (z * y)
simplify to:
(x * z) + (!x * y)
using boolean algebra?
I've been looking at the various theorems for quite awhile, but I can't figure out how this study guide I found online got to that answer.
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You can expand the term $zy$ to $xzy+x'zy$, and then $xzy$ gets absorbed by $xz$ and the term $x'zy$ gets absorbed by $x'y$. Also: this particular equivalence is called the Consensus Theorem – Bram28 Apr 06 '18 at 18:19
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If a possible transformation of one expression into the other one is not obvious, you can still blow up both expressions to their full disjunctive form:
$$xz + \bar x y + zy = x(y+\bar y)z + \bar xy(z+\bar z) + (x+\bar x)yz\\=xyz + x\bar yz + \bar xyz + \bar xy\bar z + xyz + \bar xyz\\ = xyz + \bar xyz + x\bar yz + \bar xy\bar z$$ and $$xz + \bar xy = x(y+\bar y)z + \bar xy(z+\bar z)\\ = xyz + x\bar yz + \bar xyz + \bar xy\bar z$$
As both expressions are equal, you have now a transformation of one expression into the other one.
Arthur
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trancelocation
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So other than taking a guess that (x * z) + (!x * y) is the right answer and then verifying after the fact, is there no way to get to (x * z) + (!x * y) from (x * y * z) + (!x * y * z) + (x * !y * z) + (!x * y * !z) ? – Evan Apr 10 '18 at 20:00
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You can draw a truth table and compare the output of the two expressions.
Basically for $z = y$, both expressions will output $z$ ($=y$).
For $z = 1, y = 0$, both will output $x$.
For $z = 0, y = 1$, both will output $!x$.
Deepak
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