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I've been working on this expression, but all my attempts have failed to simplify it further.

$$A'.B' + A'.B.C' + A'.B.C + A.B'.C'$$

I have tried to pick out $A'$ based on the distribution law:

$$A' (B' + B.C' + B.C) + A.B'.C'$$

Then I tried to cancel $C'$:

$$A' (C' + B.C) + A.B'.C'$$

$$A' (B) + A B' C'$$

$$A'.B + A. B'.C'$$

But this doesn't seem to be valid... any suggestions?

flonk
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dntk
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    Things have gone wrong when you try to cancel $C'$. It's true that $BC'+BC=B$, so you should be left with: $$A'(B'+B) + AB'C'$$ which, of course, simplifies further to: $$A'+AB'C'.$$ You can further simplify this to: $$A' + B'C'.$$ – Unwisdom Feb 21 '14 at 14:20

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