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Section 6.3 Direct proof
Recall.
The argument
\begin{equation*}
A \lgccond C_1, C_1 \lgccond C_2, \dotsc, C_{m-1} \lgccond C_m, C_m \lgccond B
\therefore A \lgccond B
\end{equation*}
Procedure 6.3.1 . Direct proof.
To prove \(P \lgcimplies Q \text{,}\) 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 \((\forall x)\bbrac{P(x) \lgcimplies Q(x)} \text{,}\) 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 \(P(x) \) ”, 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.
Worked Example 6.3.2 .
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.
