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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.
