Can this be simplified any further? (Boolean algebra)

43 Views Asked by At

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?