DeMorgan's Theorem

[Sathed]

I have to rebuild the following code using DeMorgan's Theorem for a class, but when I looked it up I was totally lost. Can any of you math geniuses out there lend me a hand?

if (!(today == SUN || today == SAT))
cout << "\nGo to work!" << endl;

[jsmith]

De Morgan's Theorem says that the expression

A and B

can be rewritten as

NOT ( NOT A or NOT B )

Let A be the expression ( today == SUN ) and B be the expression ( today == SAT ).

[Sathed]

When you put it like that, it's all so easy...

Thanks!