Consider the set equation (A-B) U (B-A) U (A?B)= (AUB). a) Prove that the set eq
ID: 2903056 • Letter: C
Question
Consider the set equation (A-B) U (B-A) U (A?B)= (AUB).
a) Prove that the set equation is true for all sets A and B using set operation properties (state the names of the properties in proof).
b) Prove that the set equation is true for all sets A and B usuing an element-wise proof.
c) Using the fact that set equations, logical quivalences, and Boolean equations have "equivalent" truth values, use this idea and the duality principle to write five additional statements that are also true (you must eliminate set subtractions before starting).
Explanation / Answer
The if argument is very simple. If A=B then A U B = A = B and A intersection B = A = B. Hence A U B = (A intersects B)
For the only if argument, consider an element e in A U B. e can belong to exactly one of the three disjoint sets (only A), (A intersection B) or (only B), where, (only A) denotes (AUB) ~ B and (only B) denotes (AUB) ~ A.
But it is given than (A intersection B) = A U B
Hence every element of A U B is in (A intersection B)
Hence (only A) and (only B) are null sets.
So, AUB ~ A = empty set, meaning that AUB = A
Similarly, AUB = B. Hence A = B
Btw, the argument given in the answer above mine is completely wrong.
First of all, there is no + operation in sets.
Secondly, even if the + was replaced by a U, A U B would not be {A, X} since {A, X} denotes a set of sets. Similarly, A intersection B would not be {A}.
Thirdly, to even start that argument you would require the set X to be disjoint from A, and you can't magically pull out that X from thin air. You have to first prove that you can construct such a set X given any two sets A and B.