\begin{equation*} A \rightarrow C_1, C_1 \rightarrow C_2, \dotsc, C_{m-1} \rightarrow C_m, C_m \rightarrow B \therefore A \rightarrow B \end{equation*}
is valid (Extended Law of Syllogism).
- To prove \(P \Rightarrow Q\text{,}\) start by assuming that \(P\) is true. Then, through a sequence of (appropriately justified) intermediate conclusions, arrive at \(Q\) as a final conclusion.
- To prove \((\forall x)(P(x) \Rightarrow Q(x))\text{,}\) start by assuming that \(x\) is an arbitrary but unspecified element in the domain such that \(P(x)\) is true. The first sentence in your argument should be: “Suppose \(x\) is a such that \(P(x)\)”, where the blank is filled in by the definition of the domain of \(x\text{.}\) Then, through a sequence of (appropriately justified) intermediate conclusions that do not depend on knowing the specific object \(x\) in the domain, arrive at \(Q(x)\) as a conclusion.
Prove: If \(n\) is even, then \(n^2\) even.
Solution
Let \(P(n)\) represent the predicate “\(n\) is even” and let \(Q(n)\) represent the predicate “\(n^2\) is even”, with domain the integers.
Suppose that \(n\) is an arbitrary (but unspecified) integer such that \(n\) is even. Then there exists an integer \(m\) such that \(n = 2m\text{,}\) and so \(n^2 = 4m^2 = 2(2m)\) is even.
Check your understanding. Attempt Exercises 6.12.4–6.12.6.