Exhaustive and Mutually Exclusive Case Splits

To prove that n(n+1) is even for any integer n, a common approach is to split on whether n is even or odd. If n is even, then n is divisible by 2, so n(n+1) is even. If n is odd, then n+1 is even, so n(n+1) is even. Thus, in both cases, the product is even. This split is exhaustive because every integer is either even or odd. Splitting on n being prime is not exhaustive, as non-prime integers exist. Splitting on n being positive misses negative integers, but the property holds for all integers. No split might be possible, but case analysis with even/odd is straightforward and reliable.

For natural numbers modulo 2, the possible remainders are 0 and 1. Therefore, the set {0,1} represents all cases: numbers divisible by 2 (remainder 0) and numbers not divisible by 2 (remainder 1). The set {0} only covers even numbers, missing odd numbers. {1,2} is incorrect because 2 mod 2 is 0, not 1 or 2, and it misses 0. {2} is not a valid remainder mod 2. Thus, {0,1} is exhaustive.

When splitting on x 0, the case x = 0 is not covered by either split. For real numbers, x = 0 is a distinct case that must be considered to ensure exhaustiveness. If x = 0 is not included, there might be properties that hold for positive and negative x but not for zero, leading to an incomplete proof. Therefore, x = 0 must be considered separately.

The split on “x is rational or irrational” is exhaustive for real numbers because every real number is either rational or irrational. There are no real numbers that are neither rational nor irrational. Therefore, this case split covers all real numbers, making it exhaustive. It is not incomplete, circular, or invalid based on the split alone.

Case analysis involves breaking down a complex problem into smaller, more manageable subproblems (cases) and solving each subproblem separately. This is similar to the general problem-solving strategy of division into subproblems. Direct proof proves the statement without cases, proof by contradiction uses negation, and induction is based on base cases and inductive steps. Thus, dividing a problem into subproblems is the closest similarity.

When cases are not mutually exclusive, i.e., when A∩B≠∅, elements in the overlap belong to both cases. To avoid double-counting or missing nuances, it is best to handle the overlap x∈A∩B as a separate case. This ensures that the proof covers all elements correctly and that the conclusion is established for elements in the overlap. Ignoring the overlap might lead to incomplete reasoning. Merging A and B would lose the case structure. Canceling the proof is unnecessary.

The split on “n is even” and “n is odd” is exhaustive for integers because every integer is either even or odd. This includes negative integers, zero, and positive integers. It is not limited to primes or naturals. Therefore, it is exhaustive for all integers.

For integers modulo 4, the possible remainders are 0, 1, 2, and 3. Therefore, the set {0,1,2,3} includes all possible cases. The set {0,1,2} misses remainder 3, {1,2,3} misses remainder 0, and {0,2} misses 1 and 3. Thus, {0,1,2,3} is exhaustive.

In each case of a case analysis, the sub-proof should assume the hypothesis for that case. For example, if the overall hypothesis is that P or Q is true, in the case where P is true, we assume P and prove the conclusion. Similarly for Q. Ignoring the hypothesis would not leverage the case assumptions. Using a different hypothesis or proving a different conclusion would not contribute to the overall proof. Therefore, assuming the hypothesis is correct.

The statement “If x² > 0, then x≠0,” can be proven by case analysis on the sign of x. If x>0, then x≠0 is true. If x0 is false, and the implication is vacuously true. However, to prove directly, splitting on x>0 or x0 since x²>0 implies x>0 or x

To prove that for any integer n, n² mod 3 is 0 or 1, we can split based on n mod 3. The possible remainders are 0,1,2. If n mod 3 = 0, then n² mod 3 = 0. If n mod 3 = 1, then n² mod 3 = 1. If n mod 3 = 2, then n² mod 3 = 4 mod 3 = 1. Thus, in all cases, n² mod 3 is 0 or 1. Splitting on even or odd is not sufficient because it doesn't cover mod 3 properties. Splitting on prime is not exhaustive. Therefore, n mod 3 = 0,1,2 is valid and exhaustive.

Case analysis is a flexible method that can be combined with other proof techniques such as direct proof, contradiction, induction, etc. Within each case, we might use a direct proof, or for some cases, we might use contradiction. There is no restriction to using only one method. Therefore, it can be combined with other proof techniques.

For integers modulo 5, the possible remainders are 0,1,2,3,4. Therefore, the set {0,1,2,3,4} includes all cases. The sets {0,1,2,3} and {1,2,3,4} miss remainder 4 and 0 respectively, and {0,2,4} misses 1 and 3. Thus, {0,1,2,3,4} is exhaustive.

The expression n² + n = n(n+1) is always even for integers n. The standard proof uses case analysis on n mod 2, i.e., whether n is even or odd. If n is even, then n(n+1) is even. If n is odd, then n+1 is even, so n(n+1) is even. This split is exhaustive. Splitting on n mod 3 or n being prime is not necessary or exhaustive for this property. Splitting on n>0 misses negative integers, but the property holds for all integers.

The key requirement for case analysis is that every element in the domain falls into at least one case, i.e., the cases are exhaustive. This ensures that no element is left unconsidered. It is not sufficient that at least one case works; all cases must work for all elements. Cases need not be interesting or complicated; they should be chosen to cover the domain logically.