Then there is a 1:1 mapping between the Natural numbers and the Even numbers.

Proof. Consider the map f(n) := 2n. Then for every integer n, there exists an even integer f(n). Consider two distinct integers j and k, then f(j) = 2j and f(k) = 2k, and since j does not equal k, we know that 2j does not equal 2k. Thus, f is a 1:1 map.

Wanted to try a hand-written proof, image quality is not very good..

Not quite right -- got stuck at the end and realized I had written the conjecture down incorrectly, but I wanted to post it anyways because I was pleased that I figured out what I had done wrong. Will take a second stab at it another day.

Let A, B, and C be sets. Then A \ ( B U C) = (A \ B) U (A \ C).

Proof. We show this with containment in both directions. Consider x in A \ (B U C), then x in A, but x not in B U C. This means x not in B and x not in C. Since x in A and x not in B, then x in (A \ B). Since x in A but x not in C, then x in (A \ C). Since x in (A\B) and x in (A\C), then x in (A\B) U (A \C). Thus, A \ (B U C) is a subset of (A\B) U (A\C).

To show the reverse, consider y in (A\B) U (A\C). The either y in A\B or y in A\C or both. If y in A\B then y in A and y not in B. If y in A\C, then y in A and y not in C. Thus, y in A and either y not in B or y not in C or both. ?