Finding the atoms of a Boolean Algebra

Question

I have a homework question that asks me to find the atoms of the Boolean Algebra that contains 256 Boolean functions "such as F1(x,y,z) = x + y +z, F2(x,y,z) = x + xz, F3(x,y,z) = xyz+ xyz and so on". He gives an example in his notes that for B^2, that the atoms are the elements (0,1) and (1,0). From what I understand, the homework question is asking, essentially, for the atoms of B^3.

Would the answer for this be (1,0,0), (0,1,0), (0,0,1), (1,1,0) and so on, giving me 6 atoms?

I'm not sure if I'm over simplifying my answer.

Here is the question in it's entirety:

Thanks.

Answer 1

No, you only need (1,0,0), (0,1,0), and (0,0,1) to get all the others.

For example, something like (1,1,0) can be constructed by (1,0,0)+(0,1,0), so (1,1,0) is non-atomic. Note that (0,0,0) can be formed by (1,0,0)*(0,1,0), so (0,0,0) is non-atomic as well.