Simplifying Boolean Algebra: How to Simplify Complex Boolean Expressions

AI Thread Summary
The discussion revolves around simplifying the Boolean expression (A OR C) AND NOT(C AND A AND B OR C AND A AND NOT B). Participants express confusion about how to correctly apply Boolean algebra rules, particularly the use of negation and distribution. A key point of contention is the application of the rule (A+B)' = A'B' in the context of the expression (CAB + CAB')'. Clarifications are made regarding the importance of handling each step methodically to avoid errors. Ultimately, the original poster realizes their mistake and acknowledges the need for a more deliberate approach to problem-solving in Boolean algebra.
Jaehyun
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Homework Statement
(A OR C) AND NOT(C AND A AND B OR C AND A AND NOT B)
or
(A + C) (CAB + CAB')'

Relevant Equations
(A+B)' = A'B'
A(B+C) = (AB) + (AC)
(AB)' = A' + B'

The attempt at a solution
I'm not sure how I'm suppose to expand (CAB + CAB')' for simplifying. I keep arriving at false which shouldn't be the case.

(A + C) (CA)' (B + B')' (I'm not sure if this is what your suppose to do)

or

(A + C) (C'A'B'C'A'B) (Not sure if i used (A+B)' = A'B' correctly)

Thanks - Jay
 
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Hello Jaehyun, :welcome:

Pity you deleted part of the template: relevant equations are needed to do what you want. List a few and you'll see which you need
 
BvU said:
Hello Jaehyun, :welcome:

Pity you deleted part of the template: relevant equations are needed to do what you want. List a few and you'll see which you need

I'm confused on how this rule: (A+B)' = A'B' is used to help with (CAB + CAB')'.
 
I was searching for (xyp + xyq) = xy (p+q)
 
BvU said:
I was searching for (xyp + xyq) = xy (p+q)

I don't think this rule would work as it leads to a false (CA)' (B + B')' = (CA)' (1)' = 0 unless this does not work for NOT.
 
No. Do one step at a time.
 
By the way, (XY)' ≠ X'Y' !
 
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Likes Jaehyun
(A + C) (CAB + CAB')'
(A + C) (CA)' (B + B')'
(A + C) (C' + A') (B + B')'
(AC' + AA' + CC' + CA') (B + B')'
(AC' + CA') (B + B')'

So then (B + B')' = 0 meaning there is no simplified expression?
 
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Jaehyun said:
I'm confused on how this rule: (A+B)' = A'B' is used to help with (CAB + CAB')'.
My "I was searching for (xyp + xyq) = xy (p+q)" : was meant to lure you into (CAB + CAB') = CA(B+B') = CA

(A + C) (CAB + CAB')' does NOT lead to (A + C) (CA)' (B + B')' !
 
  • #10
BvU said:
My "I was searching for (xyp + xyq) = xy (p+q)" : was meant to lure you into (CAB + CAB') = CA(B+B') = CA

(A + C) (CAB + CAB')' does NOT lead to (A + C) (CA)' (B + B')' !

But (CAB + CAB') = CA(B+B') = CA is missing the NOT portion (CAB + CAB')'.

Edit: I was rushing the question so much and I finally realized what I was doing wrong thanks BvU for putting up with me it's 2am where I live and I'm clearly not in the right mind at the moment. Solved.
 
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  • #11
Yes. The NOT portion comes afterwards. (CA)' is easier to do than what you had before.

Great. Advice: go a bit slower :smile: The speed will come with experience; in the beginning taking small steps and doing it right are more important.
 

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