Proof of exhaustion must show that a statement is true for any inevitable configuration, as determined by Grand Valley State University. Therefore, we must choose wisely how to divide our problem into parts and make sure that we consider all possible cases. Otherwise, the evidence is futile. Case proof is a valid argument in types of logic dealing with $lor$ disjunctions. Let $phi lor psi$ be a well-formed formula in an array proof whose main connector is the disjunction operator. The idea of case evidence is to divide the evidence into two or more cases and prove that the claim applies in each case. In either case, you add the condition associated with that case to the fact bank only for that case. As long as the cases cover all possibilities, you have proven the claim, regardless of the actual case. Case proof is closely related to the idea of using If instructions in a computer program. Proof. Per case. There are three cases: n ≤ −1, n = 0 and n ≥ 1.

The structure, form of reasoning, and formal form of proof by example generally proceed as follows: So what is the difference between case-by-case evidence and exhaustion evidence? As with any evidence, we try to convince the public of the validity of our claim; That is why we must be as clear and concise as possible. Together, we will work through many examples to ensure that the method of exhaustion is mastered. The idea behind the method of proof is that we break down evidence into smaller, more manageable conditions and prove that the comprehensive claim applies to each case. In logic and mathematics, proof by example (sometimes called an inappropriate generalization) is a logical error in which the validity of a statement is illustrated by one or more examples or cases, rather than by a full-fledged proof. [1] [2] The examples are also valid, albeit inelegant, evidence, although it has been shown that the examples covered cover all possible cases. In mathematics, proof by example can also be used to refer to attempts to illustrate a claim by proving cases of assertion, it being understood that these cases contain key ideas that can be generalized to a full-fledged proof. [4] A mathematical statement that aims to exhaust all possibilities by dividing the problem into parts and examining each exhibit or case separately is called case proof, sometimes called exhaustion proof. These examples describe the informal version of the logical rule known as existential introduction, also known as particularization or existential generalization: also refutes (proves the negation of the same) a universal conclusion. This is used in evidence by contrast.

In general discourse, proof by example can also be used to describe an attempt to make a claim using statistically insignificant examples. In this case, it may be necessary to assess the merits of each argument individually. [3] But first, let`s look at some examples of evidence by case. In some circumstances, examples may suffice as logically valid evidence. There are times when you begin a proof by clearly stating all possible cases, and then show that each case is true by taking clear and logical actions. In other cases, you may start using direct or indirect evidence, but you will turn to using evidence on a case-by-case basis to complete your argument. In this lesson, we will look at both scenarios. Usually, we will find that most evidence of this type contains two to four cases, but there is no limit to the number of cases you can have. For example, the famous “four-color map theorem” has more than 600 cases! Don`t worry, we focus on questions with four cases or less! In some scenarios, an argument may be valid for example if it leads from a singular premise to an existential conclusion (for example, proving that a statement is true for at least one case, rather than for all cases).

For example, let $chi$ be a well-formed formula such as $paren {phi vdash chi}$, $paren {psi vdash chi}$. Case 3 (No. ≤ −1). Since n ≤ −1 and n2 ≥ 0, the sentence is unique in this case. The case proof can be expressed in natural language as follows: The conclusion $chi$ does not depend on the hypothesis $phi$ or $psi$. Suppose we assume that $phi$ is true and deduce that $chi$ must be true. Therefore, it must follow that the truth of $chi$ follows from the fact of the truth of $phi$ or $psi$. Jenn, founder of Calcworkshop®, 15+ years of experience (certified and certified teacher) We are given that $phi$ is true, or $psi$ is true, or both. (where φ ( β / α ) {displaystyle varphi (beta /alpha )} is the formula formed by replacing all free occurrences of the variable α {displaystyle alpha } in φ {displaystyle varphi } with β {displaystyle beta }.) Case 1. (n == 0). Note that 02 ≥ 0, so if n = 0, the sentence is true. Notice how this statement is structured in such a way that you lead to the idea of dividing the problem into two parts: either n > 1 or n Example #2, which includes propositional logic and predicate logic, and in particular natural deduction.

Sometimes we just need to show that the argument for a number is odd or even. In other cases, we may need to demonstrate that a claim applies to a range of numbers or inequalities. The case-based proof is called for $phi lor psi$ as follows:. And that`s exactly what you`ll learn in today`s discrete math class.