DeMorgan's laws

Question

I need some help understanding DeMorgan's Law. I have read my discrete math textbook, and looked up the formula sheet my hardware professor handed us, but still find myself struggling with it. We have a circuit we have to build converting it all to NAND gates using DeMorgan's law.

Now, if anyone can explain DeMorgan's law better to me, that would be awesome. My formula I am working on is:

W’Y’+W’Z+W’X’+WXY

Please do not provide me with the full answer, but if you want to provide me with the first step, that would be awesome.

Thank you, KeiU

Answer 1

if you want to have a circuit with NAND gates, you have to have a SOP expression. Then you can complement your expression twice as below :

((W’Y’+W’Z+W’X’+WXY)')' = 
((W’Y’)'.(W’Z)'.(W’X’)'.(WXY)')'

so you have 5 NAND gate for the expression.

Answer 2

If you are looking for a conceptual understanding of DeMorgan's Laws:

$P \land Q$ says that both $P$ and $Q$ are true

$P \lor Q$ says that at least one of $P$ and $Q$ is true

OK, so the DeMorgan's Laws are:

$\neg(P \land Q) \Leftrightarrow \neg P \lor \neg Q$

$\neg(P \lor Q) \Leftrightarrow \neg P \land \neg Q$

So, to explain them:

$\neg(P \land Q)$ is true iff

$P \land Q$ is not true iff

not both $P$ and $Q$ are true iff

at least one of $P$ and $Q$ is not true iff

at least one of $\neg P$ and $\neg Q$ is true iff

$\neg P \lor \neg Q$

Likewise:

$\neg(P \lor Q)$ is true iff

$P \lor Q$ is not true iff

not at least one of $P$ and $Q$ are true iff

both $P$ and $Q$ are not true iff

both $\neg P$ and $\neg Q$ are true iff

$\neg P \land \neg Q$

Hope that helps ...