## Peirce’s 1870 “Logic Of Relatives” • Comment 12.5

The equation $(\mathit{s}^\mathit{l})^\mathrm{w} = \mathit{s}^{\mathit{l}\mathrm{w}}$ can be verified by establishing the corresponding equation in matrices: $(\mathsf{S}^\mathsf{L})^\mathsf{W} ~=~ \mathsf{S}^{\mathsf{L}\mathsf{W}}$

If $\mathsf{A}$ and $\mathsf{B}$ are two 1-dimensional matrices over the same index set $X$ then $\mathsf{A} = \mathsf{B}$ if and only if $\mathsf{A}_x = \mathsf{B}_x$ for every $x \in X.$  Thus, a routine way to check the validity of $(\mathsf{S}^\mathsf{L})^\mathsf{W} = \mathsf{S}^{\mathsf{L}\mathsf{W}}$ is to check whether the following equation holds for arbitrary $x \in X.$ $((\mathsf{S}^\mathsf{L})^\mathsf{W})_x ~=~ (\mathsf{S}^{\mathsf{L}\mathsf{W}})_x$

Taking both ends toward the middle, we proceed as follows: $\begin{array}{*{7}{l}} ((\mathsf{S}^\mathsf{L})^\mathsf{W})_x & = & \displaystyle \prod_{p \in X} (\mathsf{S}^\mathsf{L})_{xp}^{\mathsf{W}_p} & = & \displaystyle \prod_{p \in X} (\prod_{q \in X} \mathsf{S}_{xq}^{\mathsf{L}_{qp}})^{\mathsf{W}_p} & = & \displaystyle \prod_{p \in X} \prod_{q \in X} \mathsf{S}_{xq}^{\mathsf{L}_{qp}\mathsf{W}_p} \\[36px] (\mathsf{S}^{\mathsf{L}\mathsf{W}})_x & = & \displaystyle \prod_{q \in X} \mathsf{S}_{xq}^{(\mathsf{L}\mathsf{W})_q} & = & \displaystyle \prod_{q \in X} \mathsf{S}_{xq}^{\sum_{p \in X} \mathsf{L}_{qp} \mathsf{W}_p} & = & \displaystyle \prod_{q \in X} \prod_{p \in X} \mathsf{S}_{xq}^{\mathsf{L}_{qp} \mathsf{W}_p} \end{array}$

The products commute, so the equation holds.  In essence, the matrix identity turns on the fact that the law of exponents $(a^b)^c = a^{bc}$ in ordinary arithmetic holds when the values $a, b, c$ are restricted to the boolean domain $\mathbb{B} = \{ 0, 1 \}.$  Interpreted as a logical statement, the law of exponents $(a^b)^c = a^{bc}$ amounts to a theorem of propositional calculus that is otherwise expressed in the following ways: $\begin{matrix} (a \Leftarrow b) \Leftarrow c & = & a \Leftarrow b \land c \\[8pt] c \Rightarrow (b \Rightarrow a) & = & c \land b \Rightarrow a \end{matrix}$

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