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The Symmetric Difference Is Associative 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 2 (A∆B)∆C by hypothesis. The reason we are starting like this is that the definition of X ⊆ Y is that x 2 X implies that x 2 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 2 ((A∆B) [ C) n ((A∆B) \ C) by the definition of symmetric difference. We conclude that x 2 ((A∆B) [ C) and x2 = ((A∆B) \ C) by the definition of set difference. Continuing we see that either x 2 (A∆B) or x 2 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 2 (A∆B) By the definition of symmetric difference we have x 2 (A [ B) n (A \ B). It is possible to put the reason before the fact. Thus x 2 (A [ B) and x2 = (A \ B) by the definition of set difference. By the definition of union, Either x 2 A or x 2 B. Case 1.1: x 2 A It follows that x 2 A [ (B∆C) by the definition of union. We no proceed indirectly to show that x2 = B. Assume x 2 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 2 A \ B by the definition of intersection. This contradicts the fact that x2 = A \ B that was already established under the hypothesis of case 1. The assumption must therefore be false, so x2 = B. Proceed indirectly. Assume x 2 C. Since x 2 (A∆B) by the hypothesis of case 1, the definition of intersection implies that x 2 (A∆B)\C. However, before we split into cases, we established that x2 = (A∆B) \ C. This contradiction means that the assumption that x 2 C is false, i.e., x2 = C. One more time we proceed indirectly. Assume x 2 A \ (B∆C) The definition of intersection then gives x 2 B∆C. Thus, x 2 B [ C by the definition of symmetric difference. It follows that either x 2 B or x 2 C by the definition of union. This contradicts the facts established in this subcase that x2 = B and x2 = C by logic. We conclude that the latest assumption was false, i.e., x2 = A \ (B∆C). By the definition of set difference we have x 2 (A [ (B∆C)) n (A \ (B∆C)). The definition of symmetric difference then shows that x 2 A∆(B∆C) in case 1.1. Case 1.2: x 2 B We would now need to proceed to show that x 2 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 2 A∆(B∆C). Case 2: x 2 C We would probably need to split this into subcases as well. Again we would be trying to establish that x 2 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). .
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