## Peirce’s 1870 “Logic of Relatives” • Comment 11.12

### Peirce’s 1870 “Logic of Relatives” • Comment 11.12

Since functions are special cases of dyadic relations and since the space of dyadic relations is closed under relational composition — that is, the composition of two dyadic relations is again a dyadic relation — we know the relational composition of two functions has to be a dyadic relation.  If the relational composition of two functions is necessarily a function, too, then we would be justified in speaking of functional composition and also in saying the space of functions is closed under this functional form of composition.

Just for novelty’s sake, let’s try to prove this for relations that are functional on correlates.

The task is this — We are given a pair of dyadic relations:

$\begin{array}{lll} P \subseteq X \times Y & \text{and} & Q \subseteq Y \times Z \end{array}$

The dyadic relations $P$ and $Q$ are assumed to be functional on correlates, a premiss we express as follows.

$\begin{array}{lll} P : X \gets Y & \text{and} & Q : Y \gets Z \end{array}$

We are charged with deciding whether the relational composition $P \circ Q \subseteq X \times Z$ is also functional on correlates, in symbols, whether $P \circ Q : X \gets Z.$

It always helps to begin by recalling the pertinent definitions.

For a dyadic relation $L \subseteq X \times Y,$ we have the following equivalence.

$\begin{array}{lll} L ~\text{is a function}~ L : X \gets Y & \iff & L ~\text{is}~ 1\text{-regular at}~ Y. \end{array}$

As for the definition of relational composition, it is enough to consider the coefficient of the composite relation on an arbitrary ordered pair, $i\!:\!j.$  For that we have the following formula, where the summation indicated is logical disjunction.

$(P \circ Q)_{ij} ~=~ \sum_k P_{ik} Q_{kj}$

So let’s begin.

• $P : X \gets Y,$ or the fact that $P ~\text{is}~ 1\text{-regular at}~ Y,$ means there is exactly one ordered pair $i\!:\!k \in P$ for each $k \in Y.$
• $Q : Y \gets Z,$ or the fact that $Q ~\text{is}~ 1\text{-regular at}~ Z,$ means there is exactly one ordered pair $k\!:\!j \in Q$ for each $j \in Z.$
• As a result, there is exactly one ordered pair $i\!:\!j \in P \circ Q$ for each $j \in Z,$ which means $P \circ Q ~\text{is}~ 1\text{-regular at}~ Z,$ and so we have the function $P \circ Q : X \gets Z.$

And we are done.

### Resources

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