boolean algebra simplification.

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Vaskafdt
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boolean algebra simplification.

Postby Vaskafdt » Sun Jul 22, 2012 11:37 am UTC

well, I am stumped on a homework assignment.

I need to prove that ab'+bc'=(bc)'(a+bc')

so i went from this:
(bc)'(a+bc')=(b'+c')(a+bc')=ab'+b'bc+ac'+bc'c'=ab'+ac'+bc'=?

And now I'm stumped. I know it to be true, I have tried both a truth table.. and a karnaugh map. but I tried all kinds of manipulation and I can't reach it with the algebra simplification alone.

any ideas?
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mr-mitch
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Re: boolean algebra simplification.

Postby mr-mitch » Sun Jul 22, 2012 1:05 pm UTC

ac' = a(b+b')c' = abc' + ab'c'

ab' is in ab'c', and bc' is in abc', so you can ignore this one.

In other terms,

ab' + ac' + bc' = ab'(1 + c') + bc'(1 + a) = ab' + bc'.

My thinking behind this is that you have ab' + bc' = ab' + bc' + ac', you have this extra + ac'.

So the 1s that come from + ac' must already be included in ab' + bc'. The only way to find these ones is to inflate ac' using 1 = b + b', and so on if there are other variables.

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Vaskafdt
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Re: boolean algebra simplification.

Postby Vaskafdt » Sun Jul 22, 2012 6:45 pm UTC

thanks a lot :)

and especially thanks for explaining the reasoning behind finding the solution.
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