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36   Proof of "Association," $$\scriptstyle{p}$$$$\scriptstyle{.\supset.q}$$.  Theorems equivalent to the definitions of $$\scriptstyle{p.q}$$, $$\scriptstyle{p\supset q}$$ in Principia. , $$\scriptstyle{t\mathsf{I}t|t}$$. As this first proof from a single formal premiss stands in a unique position I shall, without in any way obscuring the precise play of the symbols, expound it after a more heuristic order than is usually followed.

We start with the Prop. $$\scriptstyle{P\mathsf{I}\pi|Q}$$, and the Rule enabling us to pass from the truth of $$\scriptstyle{P}$$ to that of $$\scriptstyle{Q}$$; and we have to prove $$\scriptstyle{\pi}$$. This can only be reached through some proposition of the form $$\scriptstyle{A\mathsf{I}B|\pi}$$, where $$\scriptstyle{A}$$ is a truth of logic. The proof will thus consist in passing from $$\scriptstyle{P\mathsf{I}\pi|Q}$$ to $$\scriptstyle{A\mathsf{I}B|\pi}$$ by some permutative process.

A simple two-terms permutative law $$\scriptstyle{s|q}$$ we do not yet possess. Our Prop. yields only a roundabout three-terms permutation, $$\scriptstyle{s|q}$$, subject to the condition of $$\scriptstyle{p\mathsf{I}q|r}$$ being a truth of logic. This, however, is enough for our purpose.

In the Prop., write $$\scriptstyle{t}$$ for $$\scriptstyle{p}$$, $$\scriptstyle{q}$$, $$\scriptstyle{r}$$: $$\scriptstyle{Q_1}$$ being $$\scriptstyle{s|t}$$. Write now $$\scriptstyle{\pi}$$ for $$\scriptstyle{p}$$, $$\scriptstyle{q}$$; $$\scriptstyle{Q_1}$$ for $$\scriptstyle{r}$$: then by (a) and the Rule, From (b) in the same manner, This enables us to pass, by the Rule, from $$\scriptstyle{P\mathsf{I}\pi|Q}$$ to In order to complete the proof of $$\scriptstyle{\pi}$$, we need only find some expression which: (α) can be a value for $$\scriptstyle{P}$$, i.e. is a case of $$\scriptstyle{p\mathsf{I}q|r}$$, and (β) is implied in some truth of logic, say $$\scriptstyle{T}$$. For, by $$\scriptstyle{T\mathsf{I}P|P}$$, the Prop., and the Rule, as above, In (e), write $$\scriptstyle{Q|\pi}$$ for $$\scriptstyle{s}$$: first by (d) and the Rule, then by $$\scriptstyle{T}$$ and the Rule, we obtain $$\scriptstyle{T\mathsf{I}Q|\pi}$$, and so