http://www.swiftglobal.com/temp/truth.gif

Thanks!

Looks fine to me... it's late though :) so no guarantees.

that's all correct.

So does it make sense that !(A && B) is not the same as A || B?

Please confirm.

Yes. Your tables are correct. !(A && B) is not the same as A || B because when A and B are 1, A || B is 1. For those inputs !(A && B) is 0, so the 2 expressions are not equivalent. Does that make sense?

The key thing to remember when doing truth tables is to consider all of the inputs and take your time. As long as you cover each of the cases it is pretty tough to screw it up. One other thing to remember is that you can break complex expressions down into smaller more manageable chunks to solve. Otherwise you'll overwhelm yourself trying to try out every possibility in one table.

I hope this helps.

If you're curious, !(A && B) = !A || !B is half of DeMorgan's rule. That may be what you were thinking of.

!(A && B) is NAND
A || B is OR
!(A || B) is NOR