## Peirce’s 1870 “Logic Of Relatives” • Comment 10.11

Let us return to the point where we left off unpacking the contents of CP 3.73.  Here Peirce remarks that the comma operator can be iterated at will:

In point of fact, since a comma may be added in this way to any relative term, it may be added to one of these very relatives formed by a comma, and thus by the addition of two commas an absolute term becomes a relative of two correlates.

So:

$\mathrm{m,\!,\!b,\!r}$

interpreted like

$\mathfrak{g}\mathit{o}\mathrm{h}$

means a man that is a rich individual and is a black that is that rich individual.

But this has no other meaning than:

$\mathrm{m,\!b,\!r}$

or a man that is a black that is rich.

Thus we see that, after one comma is added, the addition of another does not change the meaning at all, so that whatever has one comma after it must be regarded as having an infinite number.

(Peirce, CP 3.73)

Again, let us check whether this makes sense on the stage of our small but dramatic model.  Let’s say that Desdemona and Othello are rich, and, among the persons of the play, only they.  With this premiss we obtain a sample of absolute terms that is sufficiently ample to work through our example:

$\begin{array}{*{15}{c}} \mathbf{1} & = & \mathrm{B} & +\!\!, & \mathrm{C} & +\!\!, & \mathrm{D} & +\!\!, & \mathrm{E} & +\!\!, & \mathrm{I} & +\!\!, & \mathrm{J} & +\!\!, & \mathrm{O} \\[6pt] \mathrm{b} & = & \mathrm{O} \\[6pt] \mathrm{m} & = & \mathrm{C} & +\!\!, & \mathrm{I} & +\!\!, & \mathrm{J} & +\!\!, & \mathrm{O} \\[6pt] \mathrm{r} & = & \mathrm{D} & +\!\!, & \mathrm{O} \end{array}$

One application of the comma operator yields the following dyadic relatives:

$\begin{array}{*{15}{c}} \mathbf{1,} & = & \mathrm{B\!:\!B} & +\!\!, & \mathrm{C\!:\!C} & +\!\!, & \mathrm{D\!:\!D} & +\!\!, & \mathrm{E\!:\!E} & +\!\!, & \mathrm{I\!:\!I} & +\!\!, & \mathrm{J\!:\!J} & +\!\!, & \mathrm{O\!:\!O} \\[6pt] \mathrm{b,} & = & \mathrm{O\!:\!O} \\[6pt] \mathrm{m,} & = & \mathrm{C\!:\!C} & +\!\!, & \mathrm{I\!:\!I} & +\!\!, & \mathrm{J\!:\!J} & +\!\!, & \mathrm{O\!:\!O} \\[6pt] \mathrm{r,} & = & \mathrm{D\!:\!D} & +\!\!, & \mathrm{O\!:\!O} \end{array}$

Another application of the comma operator generates the following triadic relatives:

$\begin{array}{*{9}{c}} \mathbf{1,\!,} & = & \mathrm{B\!:\!B\!:\!B} & +\!\!, & \mathrm{C\!:\!C\!:\!C} & +\!\!, & \mathrm{D\!:\!D\!:\!D} & +\!\!, & \mathrm{E\!:\!E\!:\!E} \\ & & & +\!\!, & \mathrm{I\!:\!I\!:\!I} & +\!\!, & \mathrm{J\!:\!J\!:\!J} & +\!\!, & \mathrm{O\!:\!O\!:\!O} \\[6pt] \mathrm{b,\!,} & = & \mathrm{O\!:\!O\!:\!O} \\[6pt] \mathrm{m,\!,} & = & \mathrm{C\!:\!C\!:\!C} & +\!\!, & \mathrm{I\!:\!I\!:\!I} & +\!\!, & \mathrm{J\!:\!J\!:\!J} & +\!\!, & \mathrm{O\!:\!O\!:\!O} \\[6pt] \mathrm{r,\!,} & = & \mathrm{D\!:\!D\!:\!D} & +\!\!, & \mathrm{O\!:\!O\!:\!O} \end{array}$

Assuming the associativity of multiplication among dyadic relatives, we may compute the product $\mathrm{m,\!b,\!r}$ by a brute force method as follows:

$\begin{array}{lll} \mathrm{m,\!b,\!r} & = & (\mathrm{C\!:\!C} ~+\!\!,~ \mathrm{I\!:\!I} ~+\!\!,~ \mathrm{J\!:\!J} ~+\!\!,~ \mathrm{O\!:\!O})(\mathrm{O\!:\!O})(\mathrm{D} ~+\!\!,~ \mathrm{O}) \\[6pt] & = & (\mathrm{C\!:\!C} ~+\!\!,~ \mathrm{I\!:\!I} ~+\!\!,~ \mathrm{J\!:\!J} ~+\!\!,~ \mathrm{O\!:\!O})(\mathrm{O}) \\[6pt] & = & \mathrm{O} \end{array}$

This says that a man that is black that is rich is Othello, which is true on the premisses of our present universe of discourse.

Following the standard associative combinations of $\mathfrak{g}\mathit{o}\mathrm{h},$ the product $\mathrm{m,\!,\!b,\!r}$ is multiplied out along the following lines, where the trinomials of the form $\mathrm{(X\!:\!Y\!:\!Z)(Y\!:\!Z)(Z)}$ are the only ones that produce a non-null result, namely, $\mathrm{(X\!:\!Y\!:\!Z)(Y\!:\!Z)(Z) = X}.$

$\begin{array}{lll} \mathrm{m,\!,\!b,\!r} & = & (\mathrm{C\!:\!C\!:\!C} ~+\!\!,~ \mathrm{I\!:\!I\!:\!I} ~+\!\!,~ \mathrm{J\!:\!J\!:\!J} ~+\!\!,~ \mathrm{O\!:\!O\!:\!O})(\mathrm{O\!:\!O})(\mathrm{D} ~+\!\!,~ \mathrm{O}) \\[6pt] & = & (\mathrm{O\!:\!O\!:\!O})(\mathrm{O\!:\!O})(\mathrm{O}) \\[6pt] & = & \mathrm{O} \end{array}$

So we have that $\mathrm{m,\!,\!b,\!r} ~=~ \mathrm{m,\!b,\!r}.$

In closing, observe that the teridentity relation has turned up again in this context, as the second comma-ing of the universal term itself:

$\begin{array}{l} \mathbf{1},\!, = \mathrm{B\!:\!B\!:\!B} ~+\!\!,~ \mathrm{C\!:\!C\!:\!C} ~+\!\!,~ \mathrm{D\!:\!D\!:\!D} ~+\!\!,~ \mathrm{E\!:\!E\!:\!E} ~+\!\!,~ \mathrm{I\!:\!I\!:\!I} ~+\!\!,~ \mathrm{J\!:\!J\!:\!J} ~+\!\!,~ \mathrm{O\!:\!O\!:\!O} \end{array}$

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