Logic Gate Help

[BruceBly]

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

 

[MikeMl]

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.

 

[DerStrom8]

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.

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!

 

[dougy83]

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