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Given:

$$x'z+yz'+x'y$$

How can I simplify it to:

$$x'z+yz'$$

I tried:

$$\begin{aligned} x'z+yz'+x'y &= x'(z+y)+y(z'+y) \\ &=x'z+(x'y+y) \\ &=x'z+y(x'+1) \\ &=x'z+y \end{aligned}$$

Are my calculations correct if so how can I reach what I wanted without using my calculations?

MarkU
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daniel
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  • Your calculation is not correct. For the `111` input it will give `1`, while the original will be `0`. Draw some truth tables – Eugene Sh. Jan 27 '21 at 18:00
  • @EugeneSh. which step isn't correct? – daniel Jan 27 '21 at 18:00
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    As was suggested, you can check each step of your work using truth tables. Make a table for the original equation, then make a table for each step. If the tables are not the same then you made a mistake in that step. We won't do your homework for you but we will give you suggestions if you ask a very specific question. – Elliot Alderson Jan 27 '21 at 18:05
  • @ElliotAlderson Hi, this isn't hw question, plus my point was not to use truth table as I want to master Boolean laws – daniel Jan 27 '21 at 18:13
  • The end of your 1st line (z' + *y*) is wrong. – brhans Jan 27 '21 at 18:21
  • The domain of the variables should definitely be added to improve the question. Such as in https://electronics.stackexchange.com/questions/519654/simplify-the-boolean-function-z-a-bar-b-barc-i-bar-a-b-barc-i-bar I initially thought the question referred to real functions and their derivatives, for which the simplification is obviously false. – G. Bergeron Jan 27 '21 at 19:07
  • @G.Bergeron The tags are disambiguating this pretty much. – Eugene Sh. Jan 27 '21 at 19:09
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    @daniel Fine, you want to master Boolean laws. The way to do that is to **find your own errors** rather than ask others to do the hard part for you. You can find your own errors if you use another method of analysis, such as a truth table. – Elliot Alderson Jan 27 '21 at 19:28
  • I don't really mind formal logic questions. If you're looking to get help on this though you should say which law you're using on each step and not skip steps to make it easy for people to proofread. – K H Jan 28 '21 at 05:39
  • @EliasNicolas You may think that, but I am genuinely trying to help the OP to learn how to solve these problems themselves. Handing out answers helps no one. Give a man a fish... – Elliot Alderson Sep 13 '21 at 20:46

1 Answers1

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The x’y term is redundant, so it can be removed—but why?

Consider just x’z + yz’. Assume this statement is true. Then either x’z or yz’ is true. Since x only appears in the first product term and y only appears in the second, we know that either x is false or y is true or both. Therefore the x’y term in the original expression is redundant since it came as a consequence of x’z + x’y’.

You can also derive the result using a Karnaugh map.

Zane Kaminski
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