I have an expression, I know the simplified answer but I would like to know the simplified answer has been arrived at. I'd like to use the example with a class I have to teach next year - thrown in right at the deep end with a topic I have never covered before ( I know, state of the education system and all taht ....)

The expression is:
Q=(A'BC)+(ABC')+(ABC)

Simplified this becomes Q = AB + BC (or even Q=B(C+A))

but which rules are used to arrive at this? How does one determine which rules are used where?

Thanks in advance!

 

Here's a list of simplification rules.

http://sandbox.mc.edu/~bennet/cs110/boolalg/rules.html

Which rule is used in each step?

Q=(A'BC)+(ABC')+(ABC)

Q=(A'BC)+(ABC')+(ABC)+(ABC)

Q=(ABC) + (A'BC) + (ABC') + (ABC)

Q=(A+A')(BC) + (AB)(C'+C)

Q=(BC) + (AB)

 

Hi,
Thanks for the pointers:

My answers:

Original:
Q=(A'BC)+(ABC')+(ABC)

? I can see that the expression has been expanded but I do not know how to get to this as being the way in ...
Q=(A'BC)+(ABC')+(ABC)+(ABC)

Rewrite the terms to facilitate applying subsequent laws. Commutative law permits this.
Q=(ABC) + (A'BC) + (ABC') + (ABC)

Associative law
Q=(A+A')(BC) + (AB)(C'+C)

Identity
Q=(BC) + (AB)

How did I do?


I did a bit of digging myself and cam up with an alternative - the answer is correct (according to the truth table) - any comment on my application?

Original:
Q=(A'BC)+(ABC')+(ABC)

BC is common in the 1st and 3rd terms, so this can be factored out:
rewrite to make clearer:
(ABC)+(A'BC)+(ABC')
Q = BC(A+A') + (ABC')

Apply identity theorem to A+A':
Q = BC + (ABC')

Factor out the B using associative property:
Q = B(C+AC')

Use absorbtion property on (C+AC') which through association is the same as C+C'A thus C+A
Q=B(C+A)

Any comments? I know that Q=B(C+A) and Q=(BC)+(AB) are equivalent from building the truth table for each, but which is the more correct answer?

Thanks

 

eagle136,

The best way to simplify a Boolean expression is to expand it out to its minterms, and then use a Karnaugh map or a tabulation method such as the Quine-McCluskey operation. The above mentioned methods systematically reduce the Boolean expression if it can be reduced. You should download the program at http://forum.allaboutcircuits.com/showthread.php?t=22347&highlight=boolean, which will give the minterms of a Boolean expression. If the minterms of the original expression and its simplification are the same, then your simplification does not have any errors.

Ratch

 

Hi Ratch,
Thanks for the reminder. I remember doing K-Maps at Uni ages ago, our class syllabus does not include them so I had thought I should steer clear, however, having spent an evening with them again (ah memories) they seem SO much simpler than tryign to work out which law/property to apply where.

One question however, the syllabus bangs on about de Morgan's laws and students will have to be able to simplify using this. Do K-Maps help here as well? Is De Morgan used to turn a SOP into POS (and vice versa) and then use the K-Map?

Thanks in advance

 

eagle136,

One question however, the syllabus bangs on about de Morgan's laws and students will have to be able to simplify using this. Do K-Maps help here as well? Is De Morgan used to turn a SOP into POS (and vice versa) and then use the K-Map?

Certainly, here's how. Using your example, Q = A'BC+ABC'+ABC ,
the minterms are 3,6,7 . The missing minterms of 0,1,2,4,5 are the complement of your expression or Q' . You can easily see that if you make a truth table. Next, plot the 5 minterms of 0,1,2,4,5 on a K-map to get a reduction of Q' = B'+A'C' . Then use DeMorgan's theorem to easily get Q = B(A+C), which is what you started out with. This was a trivial example, but it illustrates the method. Search the archives for "Boolean" to find other problems and resolutions.

This link can give you some tips.

http://forum.allaboutcircuits.com/showthread.php?t=15687&highlight=boolean

This link shows another POS to SOP conversion.

http://forum.allaboutcircuits.com/showthread.php?t=12279&highlight=boolean

Ratch