Take the 2-minute tour ×
Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It's 100% free, no registration required.

I have this boolean expression: F1 = a1'a2' + a1a2'.

Could it be reduced even more?

share|improve this question
2  
Yes. If you can't just see it, write out a logic table and see what you get. –  The Photon Dec 22 '12 at 18:55
    
looks like homework –  Erion Dec 23 '12 at 10:01
    
@Erion Not exactly. –  Billie Dec 23 '12 at 14:07
add comment

1 Answer

up vote 8 down vote accepted

Yes, factor out the a2'. You should see something interesting with what happens to a1.

If you don't, post a comment and I can help further.

When you factor out a2', you get a2'(a1' + a1)

The statement a1' + a1 means (a1 or NOT a1). It should be obvious that this is ALWAYS true and thus can be removed from the logic statement. This leaves you with F1 = a2'

In general, the tools for reducing logic expressions that I know of are algebraic manipulation (deMorgan's Laws, distributive property, etc.), Karnaugh Maps, and the Quine McCluskey Algorithm.

share|improve this answer
    
Thanks! You're right. on which law have you based? how did you know this answer? –  Billie Dec 22 '12 at 19:04
    
To be honest, I just saw it immediately... It's kind of a distributive property I guess. Let me see if I can find some more formal resources about this. –  NickHalden Dec 22 '12 at 19:05
    
Thank you. p.s. , you meant to a1'a2' + a1, am I right? –  Billie Dec 22 '12 at 19:07
    
@user1798362, sorry you are still a little off track. –  The Photon Dec 22 '12 at 19:09
    
Just think of it in words: Something is true if a2' is true and a1 is true; or its true if a2' is true and a1 is not true. So does a1 really matter to the final result? –  The Photon Dec 22 '12 at 19:10
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.