1. snick  snack Supplement: Worked Set Proofs Based on work by Meghan Allen

2. Proofs with Sets #1 version 1 Prove that for all sets A and B: A  B = A  B Proof #1: A  B = {x  U | x  ( A  B)} = {x  U | ~(x  A  x  B)} = {x  U | x  A  x  B} = {x  U | x  A  x  B} = A  B Def’n of  De Morgan’s Def’n of Def’n of  Def’n of

3. Proofs with Sets #1 version 2 Remember that for any two sets C and D, C = D iff C is a subset of D and D is a subset of C. C = D  [C  D  D  C]

4. Proofs with Sets #1 version 2 a)Prove that: A  B  A  B Pick an arbitrary x  A  B, Then x  A  B. ~(x  A  x  B) x  A  x  B x  A  x  B x  ( A  B) Def’n of  De Morgan’s Def’n of Def’n of 

5. Proofs with Sets #1 version 2 b)Prove that: A  B  A  B Pick an arbitrary x  A  B Then, x  A  x  B x  A  x  B ~(x  A  x  B) x  A  B x  A  B

6. Proofs with Sets #1 version 2 conclusion We have shown that A  B  A  B and A  B  A  B. Therefore, A  B = A  B.

7. Proof with Sets #2 Prove that, for all sets A, B and C: (A  B) – C = (A – C)  (B – C) This time we’ll use the Set Identities LHS: (A  B) – C = (A  B)  C Set Difference Law = C  (A  B) Commutative Law = (C  A)  (C  B) Distributive Law = (A  C)  (B  C) Commutative Law = (A - C)  (B – C) Set Difference Law

8. Proofs with Sets #3 Prove or disprove, for all sets A and B, A – B = B – A We can disprove this with a counterexample. Let A = {1,2} and B = {2,3} Then, A-B = {1} and B-A = {3} A – B ≠ B – A

9. Proofs with Sets #4 Prove that for all sets A and B, if A  B then A  B =  Proof by contradiction. Assume A  B and A  B  . Then, there exists an element x  A  B. By the definition of intersection, that means x  A and x  B. Since x  B, x  B by the definition of complement. But, A  B, so since x  A, x  B by definition of subset. Thus, x  B and x  B which is a contradiction. Hence the assumption is false and therefore for all sets A and B, if A  B then A  B = 

10. Proofs with Sets #5 Prove that for all sets A and B, if A  B then P(A)  P(B) WLOG, let A and B be sets. Proof by antecedent assumption: Assume A  B. Now, consider X  P(A). X  A by the definition of power set. But, because A  B, X  B by the transitive property of subsets* and thus, by the definition of power set, X  P(B). This proves that for all X, if X  P(A) then X  P(B) and so P(A)  P(B). *Epp -Theorem 5.2.1

11. Proofs with Sets #6 version 1 Prove that for all sets S,   S. We proceed by contradiction. Assume that there is a set S such that the empty set is not a subset of S. Then, by definition of subset, there must be some element x of the empty set that is not an element of S. However, the empty set has no elements. This is a contradiction. QED

12. Proofs with Sets #6 version 2 Prove that for all sets S,   S. Note that   S   x  U, x    x  S   x  U, x    x  S. We proceed by proving this last statement. We know nothing is an element of the empty set:  x  U, x  . By generalization, it follows that:  x  U, x    x  S. QED