Differential Logic • 17

Enlargement and Difference Maps

Continuing with the example $pq : X \to \mathbb{B},$ the following venn diagram shows the enlargement or shift map $\mathrm{E}(pq) : \mathrm{E}X \to \mathbb{B}$ in the same style of field picture we drew for the tacit extension $\boldsymbol\varepsilon (pq) : \mathrm{E}X \to \mathbb{B}.$

$\text{Enlargement}~ \mathrm{E}(pq) : \mathrm{E}X \to \mathbb{B}$

$\begin{array}{rcccccc} \mathrm{E}(pq) & = & p & \cdot & q & \cdot & \texttt{(} \mathrm{d}p \texttt{)(} \mathrm{d}q \texttt{)} \\[4pt] & + & p & \cdot & \texttt{(} q \texttt{)} & \cdot & \texttt{(} \mathrm{d}p \texttt{)~} \mathrm{d}q \texttt{~} \\[4pt] & + & \texttt{(} p \texttt{)} & \cdot & q & \cdot & \texttt{~} \mathrm{d}p \texttt{~(} \mathrm{d}q \texttt{)} \\[4pt] & + & \texttt{(} p \texttt{)} & \cdot & \texttt{(} q \texttt{)} & \cdot & \texttt{~} \mathrm{d}p \texttt{~~} \mathrm{d}q \texttt{~} \end{array}$

A very important conceptual transition has just occurred here, almost tacitly, as it were. Generally speaking, having a set of mathematical objects of compatible types, in this case the two differential fields $\boldsymbol\varepsilon f$ and $\mathrm{E}f,$ both of the type $\mathrm{E}X \to \mathbb{B},$ is very useful, because it allows us to consider those fields as integral mathematical objects which can be operated on and combined in the ways we usually associate with algebras.

In the present case one notices the tacit extension $\boldsymbol\varepsilon f$ and the enlargement $\mathrm{E}f$ are in a sense dual to each other. The tacit extension $\boldsymbol\varepsilon f$ indicates all the arrows out of the region where $f$ is true and the enlargement $\mathrm{E}f$ indicates all the arrows into the region where $f$ is true. The only arc they have in common is the no‑change loop $\texttt{(} \mathrm{d}p \texttt{)(} \mathrm{d}q \texttt{)}$ at $pq.$ If we add the two sets of arcs in mod 2 fashion then the loop of multiplicity 2 zeroes out, leaving the 6 arrows of $\mathrm{D}(pq) = \boldsymbol\varepsilon(pq) + \mathrm{E}(pq)$ shown in the following venn diagram.

$\text{Difference}~ \mathrm{D}(pq) : \mathrm{E}X \to \mathbb{B}$

$\begin{array}{rcccccc} \mathrm{D}(pq) & = & p & \cdot & q & \cdot & \texttt{((} \mathrm{d}p \texttt{)(} \mathrm{d}q \texttt{))} \\[4pt] & + & p & \cdot & \texttt{(} q \texttt{)} & \cdot & \texttt{~(} \mathrm{d}p \texttt{)~} \mathrm{d}q \texttt{~~} \\[4pt] & + & \texttt{(} p \texttt{)} & \cdot & q & \cdot & \texttt{~~} \mathrm{d}p \texttt{~(} \mathrm{d}q \texttt{)~} \\[4pt] & + & \texttt{(} p \texttt{)} & \cdot & \texttt{(}q \texttt{)} & \cdot & \texttt{~~} \mathrm{d}p \texttt{~~} \mathrm{d}q \texttt{~~} \end{array}$