First, if some of these statements is false, then there is a counterexample where w has only two subsets. Also, if a statement is true, the proof is basically the same for any number of elements in w, including infinity, as long as this number is greater than one. So, it is sufficient to consider regular binary unions and intersections.

Second, if a statement is false, that would be for a non-injective F. So, you can consider, say, F(x) = x^2 and the two subsets {x | x < 0} and {x | x > 0} to see if it is a counterexample. If a statement is true, then it can be proved in the usual way, by showing that the two sides of the equality are subsets of each other.

Here is a proof that . I'll write to denote "there exists an x such that P(x)" (the scope of the quantifier extends as far to the right as possible), to denote "and", and to denote "or". Note that is equivalent to . Also, implies , but the converse is false in general (can you think of a counterexample?).

So, <=> <=> . The last statement implies , which is equivalent to and to , but what about the converse implication? Well, if , then both y's whose existence is claimed must be the same because they are F(x) and F is a function. Therefore, there exists a single y that makes both halves true, i.e., .