Boolean expression needs simplifying

  • #1
I have attached the problem and my working. It would be much appreciated if someone could check through my working and let me know if I have slipped up anywhere.

Thanks in advance

Steve
 

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Answers and Replies

  • #2
lewando
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Attachment not attached.
 
  • #3
I realised that I had made a couple of errors so I deleted the attachment. I have now reattached it above.

Also, I’ve just realised that A’+A=1… (the 0 I have put is incorrect)

So
A’+(AC) = (A’+A)+(A’+C) = 1 + (A’+C) = 1
B(1) = B

C(AB’+AC’+B) = AB’C+ACC’+BC = AB’C+BC = AB’C+BC (which somehow is the same answer as before)


Cheers

Steve
 
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  • #4
lewando
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4th line:
~A+(AC) = (~A+A)(~A+C) = 1(~A+C) = (~A+C) --got lucky.

last line:
BC(1+~A) != BC
 
  • #5
ok I see,

1 + (A’+C) is actually 1(~A+C)

Thanks for your help lewando.
 
  • #6
NascentOxygen
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What is your final simplification of the expression, Steve Collins?
 
  • #7
ac+bc

I've been staring at this one for a while:

~a~bc+~ab~c+a~b~c+abc

Can this be simplified at all?
 
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  • #8
NascentOxygen
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ac+bc
Is this your answer to my question? If so, write it as c(a+b), though that is still wrong. The easiest way to check is to draw up two truth-tables, one for the original expression, and another for your simplification. They should agree.

Please explain what rule you used to go from the LHS to the RHS here:
A’+(AC) = (A’+A)+(A’+C)

I've been staring at this one for a while:

~a~bc+~ab~c+a~b~c+abc

Can this be simplified at all?
It can be simplified using exclusive-OR operations.
 
  • #9
I've been keeping my results in sum of products form as I find it easier to read when simulating in Proteus.

A'+(AC)=(A'+A)+(A'+C) Should have read A'+(AC)=(A'+A)(A'+C)

I have constructed a truth table for the original problem and my answer and they seem to concur.
 

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  • #10
NascentOxygen
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I have constructed a truth table for the original problem and my answer and they seem to concur.
My mistake, I overlooked a closing parenthesis in your handwriting.
 
  • #11
NascentOxygen
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It can be simplified using exclusive-OR operations.
If you need encouragement: it simplifies to entirely exclusive-ORs. :smile:

¬xy + x¬y = x⊕y
 
  • #12
Thanks for the help and sorry for the late reply… Not enough hours in the day!

With NascentOxygen’s direction I’ve discovered that this Boolean expression is for a full counter so my ultimate goal is to achieve:
a xor b xor c

I have got to my working below, but I am struggling to proceed. I think that I may be going down the wrong path or there is a rule that I cannot find after much searching.

~a~bc+~ab~c+a~b~c+abc
~a(~bc+b~c)+a(~b~c+bc)
~a(b xor c)+a~(b xor c)
 
  • #13
NascentOxygen
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my ultimate goal is to achieve:
a xor b xor c

I have got to my working below, but I am struggling to proceed. I think that I may be going down the wrong path or there is a rule that I cannot find after much searching.

~a~bc+~ab~c+a~b~c+abc
~a(~bc+b~c)+a(~b~c+bc)
~a(b xor c)+a~(b xor c)
If you are able to get this far, then you are almost done.

Your last expression matches ~x.y + x.~y
 
  • #14
Thank you NascentOxygen... Your help has been greatly appreciated.
 

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