I in simplifying this boolean expression

[elorabees]

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

 

[taper100]

You could try to solve this problem using Karnaugh maps. Here is a google result I found: http://sce.umkc.edu/~hieberm/281_new/lectures/simplification/simplification.html. Example 1 may be of help.

 

[elorabees]

You could try to solve this problem using Karnaugh maps. Here is a google result I found: http://sce.umkc.edu/~hieberm/281_new/lectures/simplification/simplification.html. Example 1 may be of help.

Thank you! I'll try to solve it with that.

 

[NascentOxygen]

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 ...

 

[rezeew]

Simplifying boolean expressions involves using logical rules and laws to reduce the expression to its simplest form. In this case, we can use the following rules:

1. De Morgan's Laws: (A+B)' = A'B' and (A*B)' = A'+B'
2. Distributive Law: A*(B+C) = A*B+A*C

Using these rules, we can simplify the expression as follows:

x'y'+xz'+yz+x'yz'+xy'z
= x'y'+yz+x'yz'+xy'z (using De Morgan's Law)
= x'y'+x'yz'+xy'z+yz (rearranging terms)
= x'y'(1+z')+yz(1+x') (using Distributive Law)
= x'y'+yz (since 1+z' = 1 and 1+x' = 1)
= y(x'+z) (using Distributive Law)
= y (since x'+z = 1)

Therefore, the simplified expression is y. I hope this helps in understanding how to simplify boolean expressions.