Boolean Algebra help Please

  • Thread starter bruynz
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  • #1
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Main Question or Discussion Point

Digital Logic is blowing my head off! I can't understand a thing.

I had a question in which I was supposed to simplify the function

Code:
F= A'+B'+(A+B).B'.C
apperently this simplify to 1.

Here's what I managed to do:

Code:
F= A'+B'+[(A.B')+(B.B')].C by Distribution

F= A'+B'+(A.B').C by Tautology

F= A'+B'+A.B'.C by Distribution
according to the answer the next step get

F = A+A'+B'

F=1

however I have no idea what he's done or how to get there.

Any help would be appreciated

Regards,
Bruynz.

I hope this is the right section
 
Last edited:

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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Welcome to PF!

Hi Bruynz! Welcome to PF! :smile:
Digital Logic is blowing my head off! I can't understand a thing.

I had a question in which I was supposed to simplify the function

Code:
F= A'+B'+(A+B).B'.C
Do not adjust your mind …

reality is at fault! :biggrin:

The question is obviously wrong. :rolleyes:
 
  • #3
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F= A'+B'+(A+B).B'.C
F = A' + B' + A.B'.C
F = A' + B' + A.C
F = A' + B' + C

Yeah, they did something jive near the last step. I'm pretty sure that what I have down is the final answer. You can go back to the original and construct its truth table to trivially show it is no tautology.
 

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