Boolean expression needs simplifying

[Steve Collins]

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

 

[lewando]

Attachment not attached.

 

[Steve Collins]

I realized that I had made a couple of errors so I deleted the attachment. I have now reattached it above.

Also, I’ve just realized 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

 

[lewando]

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

last line:
BC(1+~A) != BC

 

[Steve Collins]

ok I see,

1 + (A’+C) is actually 1(~A+C)

Thanks for your help lewando.

 

[NascentOxygen]

What is your final simplification of the expression, Steve Collins?

 

[Steve Collins]

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?

 

[NascentOxygen]

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.

 

[Steve Collins]

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.

 

[NascentOxygen]

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.

 

[NascentOxygen]

It can be simplified using exclusive-OR operations.

If you need encouragement: it simplifies to entirely exclusive-ORs.

¬xy + x¬y = x⊕y

 

[Steve Collins]

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)

 

[NascentOxygen]

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

 

[Steve Collins]

Thank you NascentOxygen... Your help has been greatly appreciated.