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I have been asked to simplify the following expression in an assignment problem:

X’YZ + XYZ + XY’Z’+XYZ’ + XY’Z’ + X’Y’Z’ + WX’Y’Z’ + W’X’Y’Z’

The result of the simplification should only be 2 terms. I have attempted many times and I have been only arriving at three terms at best. Below are my steps:

(X’ + X) YZ + (Y’+Y) XZ’ + (X + X’) Y’Z’ + (W + W’) X’Y’Z’

YZ + XZ’ + Y’Z’ + X’Y’Z’

YZ + (X + Y’) Z’ + X’Y’Z’

YZ + (X + Y’ + X’Y’) Z’

YZ + (X + Y’ (1 + X’)) Z’

YZ + (X + Y’) Z’

YZ + XZ’ + Y’Z’

I am not sure how to go 2 terms from here. Any help would be much appreciated.

Thanks in advance!

Clueless Coder
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  • You have repeated the term : $XY'Z'$ Are you sure that is correct? – sai-kartik May 30 '20 at 17:04
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    Yes. Below is the question word by word:

    Simplify the given Boolean expression using the laws of Boolean logic. Show each simplification you make and please include the name of the law you used (e.g. complementation, idempotent, etc.) If you’re having trouble getting started, remember this is just like Algebra. Collect like terms and factor them out of the logical expression the exact same way you would an algebraic equation. As a hint, although it isn’t clear at first glance, the simplified expression contains just 2 terms! X’YZ + XYZ + XY’Z’+XYZ’ + XY’Z’ + X’Y’Z’ + WX’Y’Z’ + W’X’Y’Z’

     – Clueless Coder May 30 '20 at 17:09
  • thanks for confirming! I too am (unfortunately) getting 3 terms.. – sai-kartik May 30 '20 at 17:23
  • @CluelessCoder your result is correct, there was probably a typo in the given expression. – user376343 May 30 '20 at 21:26

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