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So using K-maps I was able to simplify xy + (!x)z + yz to xy + (!x)z, and I double-checked that the truth tables are the same.

I'm having trouble understanding how I would have used boolean algebra to get this result. I tried my usual tricks of adding zero, letting addition distribute over multiplication, etc., without avail.

Any thoughts appreciated.

Thanks.

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$$ xy + \bar{x} z + yz = xy + \bar{x} z + (x+\bar{x})yz = xy + \bar{x} z + xyz + \bar{x}yz = xy(1 + z) + \bar{x} z(1 + y) = xy + \bar{x}z. $$

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  • $\begingroup$ Ah okay. I had been adding 0, should have also tried multiplying by 1 like you did... thanks. $\endgroup$
    – user49404
    May 4 '21 at 18:31
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The Problem can be solved by Properties of Boolean Algebra or we also can use Consensus Theorem (also known as Redundancy Theorem in many literature)

Using Properties of Boolean Algebra
$xy+x'z+yz$
$xy+x'z+1.yz$
$xy+x'z+(x+x')yz$
$xy+x'z+xyz+xyz'$
$xy+xyz'+x'z+xyz$
$xy(1+z')+x'z+xyz$
$xy.1+x'z+xyz$
$xy+xyz+x'z$
$xy(1+z)+x'z$
$xy.1+x'z$
$xy+x'z$

Now, Consensus Theorem provides a neat and quick solution just by observing few things :

If, following requirements are fulfilled.

  1. Three variables are there in Expression.
  2. Each variable is repeated twice. (Either in Normal Form or in Complement Form)
  3. One variable is repeated in complemented form (say X)

Then,
Reject the term which doesn't contain X.

In given problem, $xy+x′z+yz$

  • A. Three Variables are there ($x, y$ and $z$)
  • $x$ is repeated Twice ($x$ and $x'$), $y$ is repeated twice ($y$ and $y$) and $z$ is repeated twice ($z$ and $z$)
  • $x$ is repeated in complemented form.

Therefore, reject term which doesn't contain $x$. Thus $yz$ is rejected.

Using same theorem $A'B'+AC'+B'C$ will be reduced to $A'B'+AC'$

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