## Differential Logic • 5

### Differential Expansions of Propositions

#### Worm’s Eye View

Let’s run through the initial example again, keeping an eye on the meanings of the formulas which develop along the way.  We begin with a proposition or a boolean function $f(p, q) = pq$ whose venn diagram and cactus graph are shown below.

A function like this has an abstract type and a concrete type.  The abstract type is what we invoke when we write things like $f : \mathbb{B} \times \mathbb{B} \to \mathbb{B}$ or $f : \mathbb{B}^2 \to \mathbb{B}.$  The concrete type takes into account the qualitative dimensions or “units” of the case, which can be explained as follows.

Let $P$ be the set of values $\{ \texttt{(} p \texttt{)},~ p \} ~=~ \{ \mathrm{not}~ p,~ p \} ~\cong~ \mathbb{B}.$
Let $Q$ be the set of values $\{ \texttt{(} q \texttt{)},~ q \} ~=~ \{ \mathrm{not}~ q,~ q \} ~\cong~ \mathbb{B}.$

Then interpret the usual propositions about $p, q$ as functions of the concrete type $f : P \times Q \to \mathbb{B}.$

We are going to consider various operators on these functions.  An operator $\mathrm{F}$ is a function which takes one function $f$ into another function $\mathrm{F}f.$

The first couple of operators we need to consider are logical analogues of two which play a founding role in the classical finite difference calculus, namely:

The difference operator $\Delta,$ written here as $\mathrm{D}.$
The enlargement operator, written here as $\mathrm{E}.$

These days, $\mathrm{E}$ is more often called the shift operator.

In order to describe the universe in which these operators operate, it is necessary to enlarge the original universe of discourse.  Starting from the initial space $X = P \times Q,$ its (first order) differential extension $\mathrm{E}X$ is constructed according to the following specifications:

$\begin{array}{rcc} \mathrm{E}X & = & X \times \mathrm{d}X \end{array}$

where:

$\begin{array}{rcc} X & = & P \times Q \\[4pt] \mathrm{d}X & = & \mathrm{d}P \times \mathrm{d}Q \\[4pt] \mathrm{d}P & = & \{ \texttt{(} \mathrm{d}p \texttt{)}, ~ \mathrm{d}p \} \\[4pt] \mathrm{d}Q & = & \{ \texttt{(} \mathrm{d}q \texttt{)}, ~ \mathrm{d}q \} \end{array}$

The interpretations of these new symbols can be diverse, but the easiest option for now is just to say $\mathrm{d}p$ means “change $p$” and $\mathrm{d}q$ means “change $q$”.

Drawing a venn diagram for the differential extension $\mathrm{E}X = X \times \mathrm{d}X$ requires four logical dimensions, $P, Q, \mathrm{d}P, \mathrm{d}Q,$ but it is possible to project a suggestion of what the differential features $\mathrm{d}p$ and $\mathrm{d}q$ are about on the 2-dimensional base space $X = P \times Q$ by drawing arrows crossing the boundaries of the basic circles in the venn diagram for $X,$ reading an arrow as $\mathrm{d}p$ if it crosses the boundary between $p$ and $\texttt{(} p \texttt{)}$ in either direction and reading an arrow as $\mathrm{d}q$ if it crosses the boundary between $q$ and $\texttt{(} q \texttt{)}$ in either direction, as indicated in the following figure.

Propositions are formed on differential variables, or any combination of ordinary logical variables and differential logical variables, in the same ways propositions are formed on ordinary logical variables alone.  For example, the proposition $\texttt{(} \mathrm{d}p \texttt{(} \mathrm{d}q \texttt{))}$ says the same thing as $\mathrm{d}p \Rightarrow \mathrm{d}q,$ in other words, there is no change in $p$ without a change in $q.$

Given the proposition $f(p, q)$ over the space $X = P \times Q,$ the (first order) enlargement of $f$ is the proposition $\mathrm{E}f$ over the differential extension $\mathrm{E}X$ defined by the following formula:

$\begin{matrix} \mathrm{E}f(p, q, \mathrm{d}p, \mathrm{d}q) & = & f(p + \mathrm{d}p,~ q + \mathrm{d}q) & = & f( \texttt{(} p, \mathrm{d}p \texttt{)},~ \texttt{(} q, \mathrm{d}q \texttt{)} ) \end{matrix}$

In the example $f(p, q) = pq,$ the enlargement $\mathrm{E}f$ is computed as follows:

$\begin{matrix} \mathrm{E}f(p, q, \mathrm{d}p, \mathrm{d}q) & = & (p + \mathrm{d}p)(q + \mathrm{d}q) & = & \texttt{(} p, \mathrm{d}p \texttt{)(} q, \mathrm{d}q \texttt{)} \end{matrix}$

Given the proposition $f(p, q)$ over $X = P \times Q,$ the (first order) difference of $f$ is the proposition $\mathrm{D}f$ over $\mathrm{E}X$ defined by the formula $\mathrm{D}f = \mathrm{E}f - f,$ or, written out in full:

$\begin{matrix} \mathrm{D}f(p, q, \mathrm{d}p, \mathrm{d}q) & = & f(p + \mathrm{d}p,~ q + \mathrm{d}q) - f(p, q) & = & \texttt{(} f( \texttt{(} p, \mathrm{d}p \texttt{)},~ \texttt{(} q, \mathrm{d}q \texttt{)} ),~ f(p, q) \texttt{)} \end{matrix}$

In the example $f(p, q) = pq,$ the difference $\mathrm{D}f$ is computed as follows:

$\begin{matrix} \mathrm{D}f(p, q, \mathrm{d}p, \mathrm{d}q) & = & (p + \mathrm{d}p)(q + \mathrm{d}q) - pq & = & \texttt{((} p, \mathrm{d}p \texttt{)(} q, \mathrm{d}q \texttt{)}, pq \texttt{)} \end{matrix}$

At the end of the previous section we evaluated this first order difference of conjunction $\mathrm{D}f$ at a single location in the universe of discourse, namely, at the point picked out by the singular proposition $pq,$ in terms of coordinates, at the place where $p = 1$ and $q = 1.$  This evaluation is written in the form $\mathrm{D}f|_{pq}$ or $\mathrm{D}f|_{(1, 1)},$ and we arrived at the locally applicable law which may be stated and illustrated as follows:

$f(p, q) ~=~ pq ~=~ p ~\mathrm{and}~ q \quad \Rightarrow \quad \mathrm{D}f|_{pq} ~=~ \texttt{((} \mathrm{dp} \texttt{)(} \mathrm{d}q \texttt{))} ~=~ \mathrm{d}p ~\mathrm{or}~ \mathrm{d}q$

The venn diagram shows the analysis of the inclusive disjunction $\texttt{((} \mathrm{d}p \texttt{)(} \mathrm{d}q \texttt{))}$ into the following exclusive disjunction:

$\begin{matrix} \mathrm{d}p ~\texttt{(} \mathrm{d}q \texttt{)} & + & \texttt{(} \mathrm{d}p \texttt{)}~ \mathrm{d}q & + & \mathrm{d}p ~\mathrm{d}q \end{matrix}$

The differential proposition $\texttt{((} \mathrm{d}p \texttt{)(} \mathrm{d}q \texttt{))}$ may be read as saying “change $p$ or change $q$ or both”.  And this can be recognized as just what you need to do if you happen to find yourself in the center cell and require a complete and detailed description of ways to escape it.

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