I in simplifying this boolean expression

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Discussion Overview

The discussion centers around the simplification of a boolean expression, specifically the expression x'y' + xz' + yz + x'yz' + xy'z. Participants seek methods for simplification and share approaches, including the use of Karnaugh maps.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant requests assistance in simplifying the boolean expression, expressing uncertainty about the process.
  • Another participant suggests using Karnaugh maps as a method for simplification and provides a link to a resource for further guidance.
  • A third participant reiterates the suggestion of using Karnaugh maps and expresses intent to try that method.
  • A later reply attempts to simplify the expression by factoring out common terms and discusses the implications of variable values on the simplification process.

Areas of Agreement / Disagreement

The discussion includes multiple approaches to the problem, with no consensus on a single method or final simplification. Participants are exploring different strategies without resolving the overall simplification.

Contextual Notes

Some steps in the simplification process are not fully resolved, and assumptions about variable values are made without explicit agreement on their implications.

elorabees
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Can someone please show me how to simplify this expression? I have no clue of how it's done so any help would be appreciated. Thank you!

x'y'+xz'+yz+x'yz'+xy'z
 
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elorabees said:
Can someone please show me how to simplify this expression? I have no clue of how it's done so any help would be appreciated. Thank you!

x'y'+xz'+yz+x'yz'+xy'z
Let's look at simplifying this expression: x'y' + x'yz'

Take out the common factor: x'(y' + yz')

Now, suppose y is false, then what's in the brackets evaluates as true.
Conversely, if y is true, then it's going to be the right-most term that determines the result, and this evaluates as z'

So, x'y' + x'yz' simplifies to: x'(y' + z')
Which can be written: x'y' + x'z'

Carry on simplifying pairs of terms ...
 

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