7.2: An application to logic

Theorem \(\PageIndex{1}\): Validity of the Extended Law of Syllogism

The Extended Law of Syllogism is a valid argument.

Proof

By mathematical induction.

Base case \(n=3\).

This is just the ordinary Law of Syllogism.

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.

\begin{array}{cc}
p_{1} \rightarrow p_{2} &\qquad p_{1} \rightarrow p_{2} \\
p_{2} \rightarrow p_{3} &\qquad p_{2} \rightarrow p_{3} \\
\vdots & \vdots \\
&\qquad p_{k-1} \rightarrow p_{k} \\p_{k-1} \rightarrow p_{k}
&\qquad p_{k} \rightarrow p_{k+1}\\ \hline{p_{1} \rightarrow p_{k} } &\hline \qquad p_{1} \rightarrow p_{k+1}
\end{array}

 

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 \(p_1 \rightarrow p_{k+1}\) 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 \(p_1 \rightarrow p_k\) to be true.

Consider the following syllogism.

\(\begin{aligned}
&p_{1} \rightarrow p_{k} \\
&p_{k} \rightarrow p_{k+1} \\
&\hline p_{1} \rightarrow p_{k+1}
\end{aligned}\)

Since this is valid (base case \(n=2\)) and its premises are all true, the conclusion is true.