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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.
