THE SYMMETRIC DIFFERENCE IS ASSOCIATIVE

DAVE AUCKLY

This is a sample proof of a result from set theory. The most important part of a proof is a chain of facts, each of which has a supporting reason. To make this clear in the following proof, I will put each fact in blue text and each reason in red text. A formal proof should have the following parts: (1) A declaration. This is a word or two to let the reader know that a statement to be proved is about to be given. For example, Theorem, Lemma, or Proposition. (2) A statement of the fact that is to be proved. This should be a complete sentence and not an imperative. (3) Something to indicate the start of the proof, e.g., “proof’.” (4) The guts of the proof which is a chain of facts and reasons leading to the statement that was to be proved. (5) Something to indicate the end of the proof. What do I use to mark the end of the following proof?? It is possible to include comments in a proof. Comments make it easier for the reader to read the proof, but are not necessary for the proof. The proof works just fine if all comments are left out. I will use green text for the comments in the following proof. Comments about the comments will be in cyan. Theorem 1. For any three sets A, B and C we have (A∆B)∆C = A∆(B∆C).

Proof. Since the definition of equality for two sets (X = Y ) is containment in both directions, i.e., X ⊆ Y and Y ⊆ X the proof will split into two parts. We indicate this with a comment at the start of the proof. We first show that (A∆B)∆C ⊆ A∆(B∆C). Let x ∈ (A∆B)∆C by hypothesis. The reason we are starting like this is that the definition of X ⊆ Y is that x ∈ X implies that x ∈ Y . This has the structure of an if-then statement. The if part is called the hypothesis and we are given that the hypothesis is true. Our goal is to prove the then statement, i.e., the conclusion. Thus, x ∈ ((A∆B) ∪ C) \ ((A∆B) ∩ C) by the definition of symmetric difference. We conclude that x ∈ ((A∆B) ∪ C) and x∈ / ((A∆B) ∩ C) by the definition of set difference. Continuing we see that either x ∈ (A∆B) or x ∈ C by the definition of union. After an either-or statement, the proof splits into cases with one case for each possible given or statement.

1 2 DAVE AUCKLY

Case 1: x ∈ (A∆B) By the definition of symmetric difference we have x ∈ (A ∪ B) \ (A ∩ B). It is possible to put the reason before the fact. Thus x ∈ (A ∪ B) and x∈ / (A ∩ B) by the definition of set difference. By the definition of union, Either x ∈ A or x ∈ B. Case 1.1: x ∈ A It follows that x ∈ A ∪ (B∆C) by the definition of union. We no proceed indirectly to show that x∈ / B. Assume x ∈ B. OK This is certainly weird the first time you see it. In an indirect proof we assume some statement *without* justification. After arriving at a contradiction we conclude that something must be wrong and the only thing that could be wrong is the assumption. This proves that the assumption was wrong. This implies that x ∈ A ∩ B by the definition of intersection. This contradicts the fact that x∈ / A ∩ B that was already established under the hypothesis of case 1. The assumption must therefore be false, so x∈ / B. Proceed indirectly. Assume x ∈ C. Since x ∈ (A∆B) by the hypothesis of case 1, the definition of intersection implies that x ∈ (A∆B)∩C. However, before we split into cases, we established that x∈ / (A∆B) ∩ C. This contradiction means that the assumption that x ∈ C is false, i.e., x∈ / C. One more time we proceed indirectly. Assume x ∈ A ∩ (B∆C) The definition of intersection then gives x ∈ B∆C. Thus, x ∈ B ∪ C by the definition of symmetric difference. It follows that either x ∈ B or x ∈ C by the definition of union. This contradicts the facts established in this subcase that x∈ / B and x∈ / C by logic. We conclude that the latest assumption was false, i.e., x∈ / A ∩ (B∆C). By the definition of set difference we have x ∈ (A ∪ (B∆C)) \ (A ∩ (B∆C)). The definition of symmetric difference then shows that x ∈ A∆(B∆C) in case 1.1. Case 1.2: x ∈ B We would now need to proceed to show that x ∈ A∆(B∆C) in case 1.2. After establishing some fact in both cases, or in this example subcases, we close this part of the argument as follows. In either case 1.1 or case 1.2 x ∈ A∆(B∆C). Case 2: x ∈ C We would probably need to split this into subcases as well. Again we would be trying to establish that x ∈ A∆(B∆C) under the conditions of case 2, so we could conclude that it was always true. The proof would continue... By the definition of symmetric containment( A∆B)∆C ⊆ A∆(B∆C). This is now the halfway point in the proof. It is good to include a comment to let the reader know where things stand. We now show that (A∆B)∆C ⊇ A∆(B∆C). Let... After the large second block, the proof would end as follows. Since (A∆B)∆C ⊆ A∆(B∆C) and (A∆B)∆C ⊇ A∆(B∆C), the definition of set equality gives( A∆B)∆C = A∆(B∆C).