Logic Gate Help

Can someone tell me if I am doing this wrong. Thanks

ABC+(AB)'C=X
ABC+(A'+B')C=X
ABC+A'C+B'C=X
C=X

I cant figure out your notation. Does A' mean not(A)?

If yes, then the Boolean reduction is correct. That means that the final logic gate implementation is a straight piece of wire.

MikeMl said:

I cant figure out your notation. Does A' mean not(A)?

If yes, then the Boolean reduction is correct. That means that the final logic gate implementation is a straight piece of wire.

Click to expand...

X' is a general notation for "not X", since computers can't show an inversion bar over an input letter.

Nice to see you over here, BruceBly!

Yes, your result is correct. You haven't explained how you got from the 3rd line to the 4th line though.

It's also simple to prove by setting Y = AB, therefore X = ABC + (AB)'C becomes X = YC + Y'C = C(Y + Y'), from which it's obvious that Y is irrelevant (Y+Y' = 1) and that X = C.

The other common way of showing e.g. "A-bar" is /A