CODEWITHSUNDEEP
booleanlogicboolean-logicalgebra
Remixt
Remixt

Reputation: 597

Boolean logic simplification wy'+wx'y+wxyz+wxz' am I doing this right?

I'm studying for my digital logic test tomorrow, and I'm doing some practice problems. I'm having problems with this one specifically. I think I solved it, but I'm not confident in my answer. Can anyone verify the logic and explain it? I have looked elsewhere but can't find the answer previously recorded anywhere.

wy'+wx'y+wxyz+wxz'

1. I factored wy out of wx'y and wxyz to get wy(x'+xz) which equates to wy(x'+z). Using the rule A+A'B = A+B

This changes the equation to wy'+wyx'+wyz+wxz'

2. I factored w out of wy' and wyx' to get w(y'+yx') which equates to w(y'+x'). Using the rule A+ A'B = A+B.

This changes the equation to wy'+wx'+wyz+wxz'

3. I factored w out of wy' and wyz to get w(y'+yz) which equates to w(y'+z'). Using the same rule again.

This changes the equation to wy'+wz'+wyx'+wxz'

4. I factored w out of wz' and wxz' to get w(z'+z'x) which equates to w(z') using the rule A+AB=A.

This changes the equation to wz'+wy'+wyx'

5. I factored w out of wy' and wyx' to get w(y'+yx') which equates to w(y'+x') using the rule A+A'B = A+B.

This changes the equation to wy'+wx'+wz' - This is where I am stuck, I feel like there are more steps to do, but I can't find anything.

Upvotes: 3

Views: 2914

Answers (1)

sfletche
sfletche

Reputation: 49744

You have a mistake in Step 3

you start with 

wy'+wx'+wyz+wxz' (equivalent to wy'+wyz+wx'+wxz')

you then correctly factor w out of wy' and wyz to get w(y'+yz)

but then you suggest w(y'+yz) is equal to w(y'+z') (which is incorrect)

w(y'+yz) is equal to w(y'+z)

which leaves you with wy'+wz+wyx'+wxz' (instead of wy'+wz'+wyx'+wxz')

after which the remaining logic after Step 3 will need to be adjusted

Upvotes: 3

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