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Section 7.2 An application to logic
Theorem 7.2.1 . Validity of the Extended Law of Syllogism.
Proof. By mathematical induction.
Base case \(n=3\) .
Induction step. Let
\(k \ge 3\text{.}\) Consider the
\(n = k\) version (below left) and the
\(n = k+1\) version (below right) of the
Extended Law of Syllogism .
\(p_1 \lgccond p_2\) \(p_2 \lgccond p_3\) \(\vdots \phantom{\lgccond p_n}\) \(p_{k-1} \lgccond p_k\) \(p_1 \lgccond p_k\)
\(p_1 \lgccond p_2\) \(p_2 \lgccond p_3\) \(\vdots\) \(p_{k-1} \lgccond p_k\) \(p_k \lgccond p_{k+1}\) \(p_1 \lgccond p_{k+1}\)
Assume the \(n = k\) version of the argument is valid. We want to show that the \(n = k + 1\) version is also valid. So suppose that premises of that latter version are all true. We need to show that the conclusion \(p_1 \lgccond p_{k+1}\) must then also be true.
But each premise of the \(n = k\) version is also a premise of the \(n = k + 1\) version, so we can say that we have assumed that every premise of the \(n = k\) version is true. But we have also assumed that version to be valid, so we may take its conclusion \(p_1 \lgccond p_k\) to be true.
Consider the following syllogism.
\(p_1 \lgccond p_k\) \(p_k \lgccond p_{k+1}\) \(p_1 \lgccond p_{k+1}\)
Since this is valid (base case \(n=2\) ) and its premises are all true, the conclusion is true.
