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.