I am currently working on proving these statements and , by means of subsets (I've heard that is is also referred to as, "proof by double inclusion."). I am having trouble with these sort of problems, so I'd certainly appreciate the help.
Here is my work:
Suppose that ; for to be in the set, can't be in . This means that ; however, by this fact, can be in , , or . Thus, when , is also in , meaning that
I have not commenced the second part, yet, because I feel as though there is something wrong with this, and I feel like I might have left out steps and explanations.
Okay, Plato, is this better:
Suppose that , then can't be an element of ; in order for to not be in , then it must be the case that , that is, it must be the case that x is not in at least one of the three sets. If is not in at least in one of those sets, then , that is, it is at least in of the three sets , , or . By the definition of the union, the set will possess the element , because at least one of the three sets has .
Would this show that the statement is a subset of the other?
And to prove that , would this be sufficient:
Suppose that . For to be in the set, has to be in at least one of the three sets, meaning can't be in , because if this were so, would have to then be in all three sets, falsifying our original assumption. Moreover, if is not in , we can say that .