Section 6.2 Common mathematical statements

In mathematics, we often want to prove that some statement \(P \) logically implies some other statement \(Q \text{.}\) That is, we want to prove that \(P \lgcimplies Q \) or \((\forall x)(P(x) \lgcimplies Q(x)) \text{.}\) Note that the universal form covers the common statement “all \(A \) are \(B \)”, since this can be rephrased “for all \(x \text{,}\) if \(x \) is \(A \) then \(x \) is \(B \)”.

Below are some common methods for proving \(P \lgcimplies Q \text{.}\) In the universal case \((\forall x)(P(x) \lgcimplies Q(x)) \text{,}\) the domain of \(x \) may be infinite, so we cannot prove \(P(x) \lgcimplies Q(x) \) for each specific \(x \text{,}\) one-by-one. Instead, we treat \(x \) as a fixed but arbitrary object in the domain, and try to construct an argument proving \(P(x) \lgcimplies Q(x) \) which does not depending on knowing the specific object \(x \text{.}\) So all of the methods below can also be used in the universal case.

Since a conditional \(P \lgccond Q \) is true automatically when \(P \) is false, it will be a tautology as long as we cannot have the case of \(P \) true and \(Q \) false at the same time. (See Figure 1.3.2.) Therefore, we (almost always) begin a proof by assuming that \(P \) is true, and proceed to demonstrate that \(Q \) must then also be true, based on that assumption.