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As stated in the title, I'm trying to simplify the following expression: $xy + xy'z + x'yz'$ I've only gotten as far as step 3:

$xy + xy'z + x'yz'$

$=x(y+y’z) + x’(yz’)$

$=x(y+y’z)+x(y’+z)$

But I don't know where to go from this step, I'm not sure if I'm allowe to rewrite y'z as y+z' (I'm not even sure if that would help)

secondubly
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  • You are not allowed to write $yz'$ as $y'+z$. Take $y=1$ and $z=0$ for a counterexample. – Rory Daulton Nov 01 '14 at 22:30
  • $yz' = (y'+z)'$, but that doesn't look like it helps any. Try $y + y'z = y + z$ – Dan Uznanski Nov 01 '14 at 22:32
  • @DanUznanski Do you know the specific law that your example illustrates? My teacher asks us to put it next to each step. – secondubly Nov 01 '14 at 22:56
  • I learned it as "absorption" but that's not the term Wikipedia uses for it. To build it from wikipedia's moves you need distributive + identity + complementation + idempotence + associative, or at least that's what I need for it: $y + y'z \Rightarrow yz'+yz+y'z \Rightarrow yz' + yz + yz + y'z \Rightarrow y+z$ – Dan Uznanski Nov 01 '14 at 23:07
  • @DanUznanski Thanks! I realized that after staring at my teacher's slides for a few minutes but I see it now. – secondubly Nov 01 '14 at 23:15

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After figuring things out with a Karnaugh map, we see that splitting up the $xy$ term will be helpful: \begin{align*} xy + xy'z + x'yz' &= xy(1) + xy'z + x'yz' \\ &= xy(z + z') + xy'z + x'yz' \\ &= (xyz + xyz') + xy'z + x'yz' \\ &= (xyz + xy'z) + (xyz' + x'yz') \\ &= xz(y + y') + yz'(x + x') \\ &= xz(1) + yz'(1) \\ &= xz + yz' \end{align*}

Adriano
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