# Differential logic/Introduction

 Subject classification: this is an Philosophy resource.

Author: Jon Awbrey

Differential logic is the component of logic whose object is the description of variation — for example, the aspects of change, difference, distribution, and diversity — in universes of discourse that are subject to logical description. A definition that broad naturally incorporates any study of variation by way of mathematical models, but differential logic is especially charged with the qualitative aspects of variation that pervade or precede quantitative models. To the extent that a logical inquiry makes use of a formal system, its differential component treats the principles that govern the use of a differential logical calculus, that is, a formal system with the expressive capacity to describe change and diversity in a logical universe of discourse.

A simple example of a differential logical calculus is furnished by a differential propositional calculus. A differential propositional calculus is a propositional calculus extended by a set of terms for describing aspects of change and difference, for example, processes that take place in a universe of discourse or transformations that map a source universe into a target universe. This augments ordinary propositional calculus in the same way that the differential calculus of Leibniz and Newton augments the analytic geometry of Descartes.

## Quick Overview

### Cactus Language for Propositional Logic

The development of differential logic is greatly facilitated by having a conceptually efficient calculus in place at the level of boolean-valued functions and elementary logical propositions. A calculus that is very efficient from both conceptual and computational standpoints is based on just two types of logical connectives, both of variable ${\displaystyle k\!~}$-ary scope. The formulas of this calculus map into a species of graph-theoretical structures called painted and rooted cacti (PARCs) that lend visual representation to their functional structure and smooth the path to efficient computation.

 The first kind of propositional expression is a parenthesized sequence of propositional expressions, written as ${\displaystyle {\texttt {(}}e_{1}{\texttt {,}}e_{2}{\texttt {,}}\ldots {\texttt {,}}e_{k-1}{\texttt {,}}e_{k}{\texttt {)}}\!~}$ and read to say that exactly one of the propositions ${\displaystyle e_{1},e_{2},\ldots ,e_{k-1},e_{k}\!~}$ is false, in other words, that their minimal negation is true. A clause of this form maps into a PARC structure called a lobe, in this case, one that is painted with the colors ${\displaystyle e_{1},e_{2},\ldots ,e_{k-1},e_{k}\!~}$ as shown below.
 The second kind of propositional expression is a concatenated sequence of propositional expressions, written as ${\displaystyle e_{1}\ e_{2}\ \ldots \ e_{k-1}\ e_{k}\!~}$ and read to say that all of the propositions ${\displaystyle e_{1},e_{2},\ldots ,e_{k-1},e_{k}\!~}$ are true, in other words, that their logical conjunction is true. A clause of this form maps into a PARC structure called a node, in this case, one that is painted with the colors ${\displaystyle e_{1},e_{2},\ldots ,e_{k-1},e_{k}\!~}$ as shown below.

All other propositional connectives can be obtained through combinations of these two forms. Strictly speaking, the parenthesized form is sufficient to define the concatenated form, making the latter formally dispensable, but it is convenient to maintain it as a concise way of expressing more complicated combinations of parenthesized forms. While working with expressions solely in propositional calculus, it is easiest to use plain parentheses for logical connectives. In contexts where ordinary parentheses are needed for other purposes an alternate typeface ${\displaystyle {\texttt {(}}\ldots {\texttt {)}}\!~}$ may be used for logical operators.

Table 1 collects a sample of basic propositional forms as expressed in terms of cactus language connectives.

 ${\displaystyle {\text{Graph}}\!~}$ ${\displaystyle {\text{Expression}}\!~}$ ${\displaystyle {\text{Interpretation}}\!~}$ ${\displaystyle {\text{Other Notations}}\!~}$ ${\displaystyle ~\!~}$ ${\displaystyle \mathrm {true} \!~}$ ${\displaystyle 1\!~}$ ${\displaystyle {\texttt {(}}~{\texttt {)}}\!~}$ ${\displaystyle \mathrm {false} \!~}$ ${\displaystyle 0\!~}$ ${\displaystyle a\!~}$ ${\displaystyle a\!~}$ ${\displaystyle a\!~}$ ${\displaystyle {\texttt {(}}a{\texttt {)}}\!~}$ ${\displaystyle \mathrm {not} ~a\!~}$ ${\displaystyle \lnot a\quad {\bar {a}}\quad {\tilde {a}}\quad a^{\prime }\!~}$ ${\displaystyle a~b~c\!~}$ ${\displaystyle a~\mathrm {and} ~b~\mathrm {and} ~c\!~}$ ${\displaystyle a\land b\land c\!~}$ ${\displaystyle {\texttt {((}}a{\texttt {)(}}b{\texttt {)(}}c{\texttt {))}}\!~}$ ${\displaystyle a~\mathrm {or} ~b~\mathrm {or} ~c\!~}$ ${\displaystyle a\lor b\lor c\!~}$ ${\displaystyle {\texttt {(}}a{\texttt {(}}b{\texttt {))}}\!~}$ ${\displaystyle {\begin{matrix}a~\mathrm {implies} ~b\\[6pt]\mathrm {if} ~a~\mathrm {then} ~b\end{matrix}}\!~}$ ${\displaystyle a\Rightarrow b\!~}$ ${\displaystyle {\texttt {(}}a{\texttt {,}}b{\texttt {)}}\!~}$ ${\displaystyle {\begin{matrix}a~\mathrm {not~equal~to} ~b\\[6pt]a~\mathrm {exclusive~or} ~b\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}a\neq b\\[6pt]a+b\end{matrix}}\!~}$ ${\displaystyle {\texttt {((}}a{\texttt {,}}b{\texttt {))}}\!~}$ ${\displaystyle {\begin{matrix}a~\mathrm {is~equal~to} ~b\\[6pt]a~\mathrm {if~and~only~if} ~b\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}a=b\\[6pt]a\Leftrightarrow b\end{matrix}}\!~}$ ${\displaystyle {\texttt {(}}a{\texttt {,}}b{\texttt {,}}c{\texttt {)}}\!~}$ ${\displaystyle {\begin{matrix}\mathrm {just~one~of} \\a,b,c\\\mathrm {is~false} .\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}&{\bar {a}}~b~c\\\lor &a~{\bar {b}}~c\\\lor &a~b~{\bar {c}}\end{matrix}}\!~}$ ${\displaystyle {\texttt {((}}a{\texttt {),(}}b{\texttt {),(}}c{\texttt {))}}\!~}$ ${\displaystyle {\begin{matrix}\mathrm {just~one~of} \\a,b,c\\\mathrm {is~true} .\\[6pt]\mathrm {partition~all} \\\mathrm {into} ~a,b,c.\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}&a~{\bar {b}}~{\bar {c}}\\\lor &{\bar {a}}~b~{\bar {c}}\\\lor &{\bar {a}}~{\bar {b}}~c\end{matrix}}\!~}$ ${\displaystyle {\texttt {(}}a{\texttt {,(}}b{\texttt {,}}c{\texttt {))}}\!~}$ ${\displaystyle {\begin{matrix}\mathrm {oddly~many~of} \\a,b,c\\\mathrm {are~true} .\end{matrix}}\!~}$ ${\displaystyle a+b+c\!~}$ ${\displaystyle {\begin{matrix}&a~b~c\\\lor &a~{\bar {b}}~{\bar {c}}\\\lor &{\bar {a}}~b~{\bar {c}}\\\lor &{\bar {a}}~{\bar {b}}~c\end{matrix}}\!~}$ ${\displaystyle {\texttt {(}}x{\texttt {,(}}a{\texttt {),(}}b{\texttt {),(}}c{\texttt {))}}\!~}$ ${\displaystyle {\begin{matrix}\mathrm {partition} ~x\\\mathrm {into} ~a,b,c.\\[6pt]\mathrm {genus} ~x~\mathrm {comprises} \\\mathrm {species} ~a,b,c.\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}&{\bar {x}}~{\bar {a}}~{\bar {b}}~{\bar {c}}\\\lor &x~a~{\bar {b}}~{\bar {c}}\\\lor &x~{\bar {a}}~b~{\bar {c}}\\\lor &x~{\bar {a}}~{\bar {b}}~c\end{matrix}}\!~}$

The simplest expression for logical truth is the empty word, usually denoted by ${\displaystyle {\boldsymbol {\varepsilon }}\!~}$ or ${\displaystyle \lambda \!~}$ in formal languages, where it forms the identity element for concatenation. To make it visible in context, it may be denoted by the equivalent expression ${\displaystyle {}^{\backprime \backprime }{\texttt {((}}~{\texttt {))}}{}^{\prime \prime },\!~}$ or, especially if operating in an algebraic context, by a simple ${\displaystyle {}^{\backprime \backprime }1{}^{\prime \prime }.\!~}$ Also when working in an algebraic mode, the plus sign ${\displaystyle {}^{\backprime \backprime }+{}^{\prime \prime }\!~}$ may be used for exclusive disjunction. For example, we have the following paraphrases of algebraic expressions by means of parenthesized expressions:

 ${\displaystyle {\begin{matrix}a+b&=&{\texttt {(}}a{\texttt {,}}b{\texttt {)}}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}a+b+c&=&{\texttt {(}}a{\texttt {,(}}b{\texttt {,}}c{\texttt {))}}&=&{\texttt {((}}a{\texttt {,}}b{\texttt {),}}c{\texttt {)}}\end{matrix}}\!~}$

It is important to note that the last expressions are not equivalent to the 3-place parenthesis ${\displaystyle {\texttt {(}}a{\texttt {,}}b{\texttt {,}}c{\texttt {)}}.\!~}$

### Differential Expansions of Propositions

#### Bird's Eye View

An efficient calculus for the realm of logic represented by boolean functions and elementary propositions makes it feasible to compute the finite differences and the differentials of those functions and propositions.

For example, consider a proposition of the form ${\displaystyle {}^{\backprime \backprime }\,p~\mathrm {and} ~q\,{}^{\prime \prime }\!~}$ that is graphed as two letters attached to a root node:

Written as a string, this is just the concatenation ${\displaystyle p~q\!~}$.

The proposition ${\displaystyle pq\!~}$ may be taken as a boolean function ${\displaystyle f(p,q)\!~}$ having the abstract type ${\displaystyle f:\mathbb {B} \times \mathbb {B} \to \mathbb {B} ,\!~}$ where ${\displaystyle \mathbb {B} =\{0,1\}\!~}$ is read in such a way that ${\displaystyle 0\!~}$ means ${\displaystyle \mathrm {false} \!~}$ and ${\displaystyle 1\!~}$ means ${\displaystyle \mathrm {true} .\!~}$

Imagine yourself standing in a fixed cell of the corresponding venn diagram, say, the cell where the proposition ${\displaystyle pq\!~}$ is true, as shown in the following Figure:

Now ask yourself: What is the value of the proposition ${\displaystyle pq\!~}$ at a distance of ${\displaystyle \mathrm {d} p\!~}$ and ${\displaystyle \mathrm {d} q\!~}$ from the cell ${\displaystyle pq\!~}$ where you are standing?

Don't think about it — just compute:

The cactus formula ${\displaystyle {\texttt {(p,dp)(q,dq)}}\!~}$ and its corresponding graph arise by substituting ${\displaystyle p+\mathrm {d} p\!~}$ for ${\displaystyle p\!~}$ and ${\displaystyle q+\mathrm {d} q\!~}$ for ${\displaystyle q\!~}$ in the boolean product or logical conjunction ${\displaystyle pq\!~}$ and writing the result in the two dialects of cactus syntax. This follows from the fact that the boolean sum ${\displaystyle p+\mathrm {d} p\!~}$ is equivalent to the logical operation of exclusive disjunction, which parses to a cactus graph of the following form:

Next question: What is the difference between the value of the proposition ${\displaystyle pq\!~}$ over there, at a distance of ${\displaystyle \mathrm {d} p\!~}$ and ${\displaystyle \mathrm {d} q,\!~}$ and the value of the proposition ${\displaystyle pq\!~}$ where you are standing, all expressed in the form of a general formula, of course? Here is the appropriate formulation:

There is one thing that I ought to mention at this point: Computed over ${\displaystyle \mathbb {B} ,\!~}$ plus and minus are identical operations. This will make the relation between the differential and the integral parts of the appropriate calculus slightly stranger than usual, but we will get into that later.

Last question, for now: What is the value of this expression from your current standpoint, that is, evaluated at the point where ${\displaystyle pq\!~}$ is true? Well, substituting ${\displaystyle 1\!~}$ for ${\displaystyle p\!~}$ and ${\displaystyle 1\!~}$ for ${\displaystyle q\!~}$ in the graph amounts to erasing the labels ${\displaystyle p\!~}$ and ${\displaystyle q\!,\!~}$ as shown here:

And this is equivalent to the following graph:

We have just met with the fact that the differential of the and is the or of the differentials.

 ${\displaystyle {\begin{matrix}p~\mathrm {and} ~q&\quad &{\xrightarrow {\quad \mathrm {Diff} \quad }}&\quad &\mathrm {d} p~\mathrm {or} ~\mathrm {d} q\end{matrix}}\!~}$

It will be necessary to develop a more refined analysis of that statement directly, but that is roughly the nub of it.

If the form of the above statement reminds you of De Morgan's rule, it is no accident, as differentiation and negation turn out to be closely related operations. Indeed, one can find discussions of logical difference calculus in the Boole–De Morgan correspondence and Peirce also made use of differential operators in a logical context, but the exploration of these ideas has been hampered by a number of factors, not the least of which has been the lack of a syntax that was adequate to handle the complexity of expressions that evolve.

#### Worm's Eye View

Let's run through the initial example again, this time attempting to interpret the formulas that develop at each stage along the way. We begin with a proposition or a boolean function ${\displaystyle f(p,q)=pq.\!~}$

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

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

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

We are going to consider various operators on these functions. Here, an operator ${\displaystyle \mathrm {F} \!~}$ is a function that takes one function ${\displaystyle f\!~}$ into another function ${\displaystyle \mathrm {F} f.\!~}$

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

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

These days, ${\displaystyle \mathrm {E} \!~}$ 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 ${\displaystyle X=P\times Q,\!~}$ its (first order) differential extension ${\displaystyle \mathrm {E} X\!~}$ is constructed according to the following specifications:

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

where:

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

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

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

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

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

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

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

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

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

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

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

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

We did not yet go through the trouble to interpret this (first order) difference of conjunction fully, but were happy simply to evaluate it with respect to a single location in the universe of discourse, namely, at the point picked out by the singular proposition ${\displaystyle pq,\!~}$ that is, at the place where ${\displaystyle p=1\!~}$ and ${\displaystyle q=1.\!~}$ This evaluation is written in the form ${\displaystyle \mathrm {D} f|_{pq}\!~}$ or ${\displaystyle \mathrm {D} f|_{(1,1)},\!~}$ and we arrived at the locally applicable law that is stated and illustrated as follows:

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

The picture shows the analysis of the inclusive disjunction ${\displaystyle {\texttt {((}}\mathrm {d} p{\texttt {)(}}\mathrm {d} q{\texttt {))}}\!~}$ into the following exclusive disjunction:

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

The differential proposition that results may be interpreted to say “change ${\displaystyle p\!~}$ or change ${\displaystyle q\!~}$ 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.

Last time we computed what is variously called the difference map, the difference proposition, or the local proposition ${\displaystyle \mathrm {D} f_{x}\!~}$ of the proposition ${\displaystyle f(p,q)=pq\!~}$ at the point ${\displaystyle x\!~}$ where ${\displaystyle p=1\!~}$ and ${\displaystyle q=1.\!~}$

In the universe ${\displaystyle X=P\times Q,\!~}$ the four propositions ${\displaystyle pq,~p{\texttt {(}}q{\texttt {)}},~{\texttt {(}}p{\texttt {)}}q,~{\texttt {(}}p{\texttt {)(}}q{\texttt {)}}\!~}$ that indicate the “cells”, or the smallest regions of the venn diagram, are called singular propositions. These serve as an alternative notation for naming the points ${\displaystyle (1,1),~(1,0),~(0,1),~(0,0),\!~}$ respectively.

Thus we can write ${\displaystyle \mathrm {D} f_{x}=\mathrm {D} f|x=\mathrm {D} f|(1,1)=\mathrm {D} f|pq,\!~}$ so long as we know the frame of reference in force.

In the example ${\displaystyle f(p,q)=pq,\!~}$ the value of the difference proposition ${\displaystyle \mathrm {D} f_{x}\!~}$ at each of the four points in ${\displaystyle x\in X\!~}$ may be computed in graphical fashion as shown below:

The easy way to visualize the values of these graphical expressions is just to notice the following equivalents:

Laying out the arrows on the augmented venn diagram, one gets a picture of a differential vector field.

The Figure shows the points of the extended universe ${\displaystyle \mathrm {E} X=P\times Q\times \mathrm {d} P\times \mathrm {d} Q\!~}$ that are indicated by the difference map ${\displaystyle \mathrm {D} f:\mathrm {E} X\to \mathbb {B} ,\!~}$ namely, the following six points or singular propositions::

 ${\displaystyle {\begin{array}{rcccc}1.&p&q&\mathrm {d} p&\mathrm {d} q\\2.&p&q&\mathrm {d} p&(\mathrm {d} q)\\3.&p&q&(\mathrm {d} p)&\mathrm {d} q\\4.&p&(q)&(\mathrm {d} p)&\mathrm {d} q\\5.&(p)&q&\mathrm {d} p&(\mathrm {d} q)\\6.&(p)&(q)&\mathrm {d} p&\mathrm {d} q\end{array}}\!~}$

The information borne by ${\displaystyle \mathrm {D} f\!~}$ should be clear enough from a survey of these six points — they tell you what you have to do from each point of ${\displaystyle X\!~}$ in order to change the value borne by ${\displaystyle f(p,q),\!~}$ that is, the move you have to make in order to reach a point where the value of the proposition ${\displaystyle f(p,q)\!~}$ is different from what it is where you started.

We have been studying the action of the difference operator ${\displaystyle \mathrm {D} \!~}$ on propositions of the form ${\displaystyle f:P\times Q\to \mathbb {B} ,\!~}$ as illustrated by the example ${\displaystyle f(p,q)=pq\!~}$ that is known in logic as the conjunction of ${\displaystyle p\!~}$ and ${\displaystyle q.\!~}$ The resulting difference map ${\displaystyle \mathrm {D} f\!~}$ is a (first order) differential proposition, that is, a proposition of the form ${\displaystyle \mathrm {D} f:P\times Q\times \mathrm {d} P\times \mathrm {d} Q\to \mathbb {B} .\!~}$

Abstracting from the augmented venn diagram that shows how the models or satisfying interpretations of ${\displaystyle \mathrm {D} f\!~}$ distribute over the extended universe of discourse ${\displaystyle \mathrm {E} X=P\times Q\times \mathrm {d} P\times \mathrm {d} Q,\!~}$ the difference map ${\displaystyle \mathrm {D} f\!~}$ can be represented in the form of a digraph or directed graph, one whose points are labeled with the elements of ${\displaystyle X=P\times Q\!~}$ and whose arrows are labeled with the elements of ${\displaystyle \mathrm {d} X=\mathrm {d} P\times \mathrm {d} Q,\!~}$ as shown in the following Figure.

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

Any proposition worth its salt can be analyzed from many different points of view, any one of which has the potential to reveal an unsuspected aspect of the proposition's meaning. We will encounter more and more of these alternative readings as we go.

The enlargement or shift operator ${\displaystyle \mathrm {E} \!~}$ exhibits a wealth of interesting and useful properties in its own right, so it pays to examine a few of the more salient features that play out on the surface of our initial example, ${\displaystyle f(p,q)=pq.\!~}$

A suitably generic definition of the extended universe of discourse is afforded by the following set-up:

 ${\displaystyle {\begin{array}{lccl}{\text{Let}}&X&=&X_{1}\times \ldots \times X_{k}.\\[6pt]{\text{Let}}&\mathrm {d} X&=&\mathrm {d} X_{1}\times \ldots \times \mathrm {d} X_{k}.\\[6pt]{\text{Then}}&\mathrm {E} X&=&X\times \mathrm {d} X\\[6pt]&&=&X_{1}\times \ldots \times X_{k}~\times ~\mathrm {d} X_{1}\times \ldots \times \mathrm {d} X_{k}\end{array}}\!~}$

For a proposition of the form ${\displaystyle f:X_{1}\times \ldots \times X_{k}\to \mathbb {B} ,\!~}$ the (first order) enlargement of ${\displaystyle f\!~}$ is the proposition ${\displaystyle \mathrm {E} f:\mathrm {E} X\to \mathbb {B} \!~}$ that is defined by the following equation:

 ${\displaystyle {\begin{array}{l}\mathrm {E} f(x_{1},\ldots ,x_{k},\mathrm {d} x_{1},\ldots ,\mathrm {d} x_{k})\\[6pt]=\quad f(x_{1}+\mathrm {d} x_{1},\ldots ,x_{k}+\mathrm {d} x_{k})\\[6pt]=\quad f({\texttt {(}}x_{1},\mathrm {d} x_{1}{\texttt {)}},\ldots ,{\texttt {(}}x_{k},\mathrm {d} x_{k}{\texttt {)}})\end{array}}\!~}$

The differential variables ${\displaystyle \mathrm {d} x_{j}\!~}$ are boolean variables of the same basic type as the ordinary variables ${\displaystyle x_{j}.\!~}$ Although it is conventional to distinguish the (first order) differential variables with the operative prefix “${\displaystyle \mathrm {d} \!~}$” this way of notating differential variables is entirely optional. It is their existence in particular relations to the initial variables, not their names, that defines them as differential variables.

In the example of logical conjunction, ${\displaystyle f(p,q)=pq,\!~}$ the enlargement ${\displaystyle \mathrm {E} f\!~}$ is formulated as follows:

 ${\displaystyle {\begin{array}{l}\mathrm {E} f(p,q,\mathrm {d} p,\mathrm {d} q)\\[6pt]=\quad (p+\mathrm {d} p)(q+\mathrm {d} q)\\[6pt]=\quad {\texttt {(}}p,\mathrm {d} p{\texttt {)(}}q,\mathrm {d} q{\texttt {)}}\end{array}}\!~}$

Given that this expression uses nothing more than the boolean ring operations of addition and multiplication, it is permissible to “multiply things out” in the usual manner to arrive at the following result:

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

To understand what the enlarged or shifted proposition means in logical terms, it serves to go back and analyze the above expression for ${\displaystyle \mathrm {E} f\!~}$ in the same way that we did for ${\displaystyle \mathrm {D} f.\!~}$ Toward that end, the value of ${\displaystyle \mathrm {E} f_{x}\!~}$ at each ${\displaystyle x\in X\!~}$ may be computed in graphical fashion as shown below:

Given the data that develops in this form of analysis, the disjoined ingredients can now be folded back into a boolean expansion or a disjunctive normal form (DNF) that is equivalent to the enlarged proposition ${\displaystyle \mathrm {E} f.\!~}$

 ${\displaystyle {\begin{matrix}\mathrm {E} f&=&pq\cdot \mathrm {E} f_{pq}&+&p(q)\cdot \mathrm {E} f_{p(q)}&+&(p)q\cdot \mathrm {E} f_{(p)q}&+&(p)(q)\cdot \mathrm {E} f_{(p)(q)}\end{matrix}}\!~}$

Here is a summary of the result, illustrated by means of a digraph picture, where the “no change” element ${\displaystyle (\mathrm {d} p)(\mathrm {d} q)\!~}$ is drawn as a loop at the point ${\displaystyle p~q.\!~}$

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

We may understand the enlarged proposition ${\displaystyle \mathrm {E} f\!~}$ as telling us all the different ways to reach a model of the proposition ${\displaystyle f\!~}$ from each point of the universe ${\displaystyle X.\!~}$

## Propositional Forms on Two Variables

To broaden our experience with simple examples, let us examine the sixteen functions of concrete type ${\displaystyle P\times Q\to \mathbb {B} \!~}$ and abstract type ${\displaystyle \mathbb {B} \times \mathbb {B} \to \mathbb {B} .\!~}$ A few Tables are set here that detail the actions of ${\displaystyle \mathrm {E} \!~}$ and ${\displaystyle \mathrm {D} \!~}$ on each of these functions, allowing us to view the results in several different ways.

Tables A1 and A2 show two ways of arranging the 16 boolean functions on two variables, giving equivalent expressions for each function in several different systems of notation.

 ${\displaystyle {\mathcal {L}}_{1}\!~}$ ${\displaystyle {\text{Decimal}}\!~}$ ${\displaystyle {\mathcal {L}}_{2}\!~}$ ${\displaystyle {\text{Binary}}\!~}$ ${\displaystyle {\mathcal {L}}_{3}\!~}$ ${\displaystyle {\text{Vector}}\!~}$ ${\displaystyle {\mathcal {L}}_{4}\!~}$ ${\displaystyle {\text{Cactus}}\!~}$ ${\displaystyle {\mathcal {L}}_{5}\!~}$ ${\displaystyle {\text{English}}\!~}$ ${\displaystyle {\mathcal {L}}_{6}\!~}$ ${\displaystyle {\text{Ordinary}}\!~}$ ${\displaystyle p\colon \!~}$ ${\displaystyle 1~1~0~0\!~}$ ${\displaystyle q\colon \!~}$ ${\displaystyle 1~0~1~0\!~}$ ${\displaystyle {\begin{matrix}f_{0}\\[4pt]f_{1}\\[4pt]f_{2}\\[4pt]f_{3}\\[4pt]f_{4}\\[4pt]f_{5}\\[4pt]f_{6}\\[4pt]f_{7}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{0000}\\[4pt]f_{0001}\\[4pt]f_{0010}\\[4pt]f_{0011}\\[4pt]f_{0100}\\[4pt]f_{0101}\\[4pt]f_{0110}\\[4pt]f_{0111}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}0~0~0~0\\[4pt]0~0~0~1\\[4pt]0~0~1~0\\[4pt]0~0~1~1\\[4pt]0~1~0~0\\[4pt]0~1~0~1\\[4pt]0~1~1~0\\[4pt]0~1~1~1\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(~)\\[4pt](p)(q)\\[4pt](p)~q~\\[4pt](p)~~\\[4pt]~p~(q)\\[4pt]~~(q)\\[4pt](p,~q)\\[4pt](p~~q)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}{\text{false}}\\[4pt]{\text{neither}}~p~{\text{nor}}~q\\[4pt]q~{\text{without}}~p\\[4pt]{\text{not}}~p\\[4pt]p~{\text{without}}~q\\[4pt]{\text{not}}~q\\[4pt]p~{\text{not equal to}}~q\\[4pt]{\text{not both}}~p~{\text{and}}~q\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}0\\[4pt]\lnot p\land \lnot q\\[4pt]\lnot p\land q\\[4pt]\lnot p\\[4pt]p\land \lnot q\\[4pt]\lnot q\\[4pt]p\neq q\\[4pt]\lnot p\lor \lnot q\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{8}\\[4pt]f_{9}\\[4pt]f_{10}\\[4pt]f_{11}\\[4pt]f_{12}\\[4pt]f_{13}\\[4pt]f_{14}\\[4pt]f_{15}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{1000}\\[4pt]f_{1001}\\[4pt]f_{1010}\\[4pt]f_{1011}\\[4pt]f_{1100}\\[4pt]f_{1101}\\[4pt]f_{1110}\\[4pt]f_{1111}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}1~0~0~0\\[4pt]1~0~0~1\\[4pt]1~0~1~0\\[4pt]1~0~1~1\\[4pt]1~1~0~0\\[4pt]1~1~0~1\\[4pt]1~1~1~0\\[4pt]1~1~1~1\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~~p~~q~~\\[4pt]((p,~q))\\[4pt]~~~q~~\\[4pt]~(p~(q))\\[4pt]~~p~~~\\[4pt]((p)~q)~\\[4pt]((p)(q))\\[4pt]((~))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}p~{\text{and}}~q\\[4pt]p~{\text{equal to}}~q\\[4pt]q\\[4pt]{\text{not}}~p~{\text{without}}~q\\[4pt]p\\[4pt]{\text{not}}~q~{\text{without}}~p\\[4pt]p~{\text{or}}~q\\[4pt]{\text{true}}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}p\land q\\[4pt]p=q\\[4pt]q\\[4pt]p\Rightarrow q\\[4pt]p\\[4pt]p\Leftarrow q\\[4pt]p\lor q\\[4pt]1\end{matrix}}\!~}$

 ${\displaystyle {\mathcal {L}}_{1}\!~}$ ${\displaystyle {\text{Decimal}}\!~}$ ${\displaystyle {\mathcal {L}}_{2}\!~}$ ${\displaystyle {\text{Binary}}\!~}$ ${\displaystyle {\mathcal {L}}_{3}\!~}$ ${\displaystyle {\text{Vector}}\!~}$ ${\displaystyle {\mathcal {L}}_{4}\!~}$ ${\displaystyle {\text{Cactus}}\!~}$ ${\displaystyle {\mathcal {L}}_{5}\!~}$ ${\displaystyle {\text{English}}\!~}$ ${\displaystyle {\mathcal {L}}_{6}\!~}$ ${\displaystyle {\text{Ordinary}}\!~}$ ${\displaystyle p\colon \!~}$ ${\displaystyle 1~1~0~0\!~}$ ${\displaystyle q\colon \!~}$ ${\displaystyle 1~0~1~0\!~}$ ${\displaystyle f_{0}\!~}$ ${\displaystyle f_{0000}\!~}$ ${\displaystyle 0~0~0~0\!~}$ ${\displaystyle (~)\!~}$ ${\displaystyle {\text{false}}\!~}$ ${\displaystyle 0\!~}$ ${\displaystyle {\begin{matrix}f_{1}\\[4pt]f_{2}\\[4pt]f_{4}\\[4pt]f_{8}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{0001}\\[4pt]f_{0010}\\[4pt]f_{0100}\\[4pt]f_{1000}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}0~0~0~1\\[4pt]0~0~1~0\\[4pt]0~1~0~0\\[4pt]1~0~0~0\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(p)(q)\\[4pt](p)~q~\\[4pt]~p~(q)\\[4pt]~p~~q~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}{\text{neither}}~p~{\text{nor}}~q\\[4pt]q~{\text{without}}~p\\[4pt]p~{\text{without}}~q\\[4pt]p~{\text{and}}~q\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}\lnot p\land \lnot q\\[4pt]\lnot p\land q\\[4pt]p\land \lnot q\\[4pt]p\land q\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{3}\\[4pt]f_{12}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{0011}\\[4pt]f_{1100}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}0~0~1~1\\[4pt]1~1~0~0\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(p)\\[4pt]~p~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}{\text{not}}~p\\[4pt]p\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}\lnot p\\[4pt]p\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{6}\\[4pt]f_{9}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{0110}\\[4pt]f_{1001}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}0~1~1~0\\[4pt]1~0~0~1\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~(p,~q)~\\[4pt]((p,~q))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}p~{\text{not equal to}}~q\\[4pt]p~{\text{equal to}}~q\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}p\neq q\\[4pt]p=q\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{5}\\[4pt]f_{10}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{0101}\\[4pt]f_{1010}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}0~1~0~1\\[4pt]1~0~1~0\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(q)\\[4pt]~q~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}{\text{not}}~q\\[4pt]q\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}\lnot q\\[4pt]q\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{7}\\[4pt]f_{11}\\[4pt]f_{13}\\[4pt]f_{14}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{0111}\\[4pt]f_{1011}\\[4pt]f_{1101}\\[4pt]f_{1110}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}0~1~1~1\\[4pt]1~0~1~1\\[4pt]1~1~0~1\\[4pt]1~1~1~0\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~(p~~q)~\\[4pt]~(p~(q))\\[4pt]((p)~q)~\\[4pt]((p)(q))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}{\text{not both}}~p~{\text{and}}~q\\[4pt]{\text{not}}~p~{\text{without}}~q\\[4pt]{\text{not}}~q~{\text{without}}~p\\[4pt]p~{\text{or}}~q\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}\lnot p\lor \lnot q\\[4pt]p\Rightarrow q\\[4pt]p\Leftarrow q\\[4pt]p\lor q\end{matrix}}\!~}$ ${\displaystyle f_{15}\!~}$ ${\displaystyle f_{1111}\!~}$ ${\displaystyle 1~1~1~1\!~}$ ${\displaystyle ((~))\!~}$ ${\displaystyle {\text{true}}\!~}$ ${\displaystyle 1\!~}$

### Transforms Expanded over Differential Features

The next four Tables expand the expressions of ${\displaystyle \mathrm {E} f\!~}$ and ${\displaystyle \mathrm {D} f\!~}$ in two different ways, for each of the sixteen functions. Notice that the functions are given in a different order, partitioned into seven natural classes by a group action.

 ${\displaystyle f\!~}$ ${\displaystyle \mathrm {T} _{11}f\!~}$ ${\displaystyle \mathrm {E} f|_{\mathrm {d} p~\mathrm {d} q}\!~}$ ${\displaystyle \mathrm {T} _{10}f\!~}$ ${\displaystyle \mathrm {E} f|_{\mathrm {d} p(\mathrm {d} q)}\!~}$ ${\displaystyle \mathrm {T} _{01}f\!~}$ ${\displaystyle \mathrm {E} f|_{(\mathrm {d} p)\mathrm {d} q}\!~}$ ${\displaystyle \mathrm {T} _{00}f\!~}$ ${\displaystyle \mathrm {E} f|_{(\mathrm {d} p)(\mathrm {d} q)}\!~}$ ${\displaystyle f_{0}\!~}$ ${\displaystyle (~)\!~}$ ${\displaystyle (~)\!~}$ ${\displaystyle (~)\!~}$ ${\displaystyle (~)\!~}$ ${\displaystyle (~)\!~}$ ${\displaystyle {\begin{matrix}f_{1}\\[4pt]f_{2}\\[4pt]f_{4}\\[4pt]f_{8}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(p)(q)\\[4pt](p)~q~\\[4pt]~p~(q)\\[4pt]~p~~q~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~p~~q~\\[4pt]~p~(q)\\[4pt](p)~q~\\[4pt](p)(q)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~p~(q)\\[4pt]~p~~q~\\[4pt](p)(q)\\[4pt](p)~q~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(p)~q~\\[4pt](p)(q)\\[4pt]~p~~q~\\[4pt]~p~(q)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(p)(q)\\[4pt](p)~q~\\[4pt]~p~(q)\\[4pt]~p~~q~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{3}\\[4pt]f_{12}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(p)\\[4pt]~p~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~p~\\[4pt](p)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~p~\\[4pt](p)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(p)\\[4pt]~p~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(p)\\[4pt]~p~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{6}\\[4pt]f_{9}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~(p,~q)~\\[4pt]((p,~q))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~(p,~q)~\\[4pt]((p,~q))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}((p,~q))\\[4pt]~(p,~q)~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}((p,~q))\\[4pt]~(p,~q)~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~(p,~q)~\\[4pt]((p,~q))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{5}\\[4pt]f_{10}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(q)\\[4pt]~q~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~q~\\[4pt](q)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(q)\\[4pt]~q~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~q~\\[4pt](q)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(q)\\[4pt]~q~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{7}\\[4pt]f_{11}\\[4pt]f_{13}\\[4pt]f_{14}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(~p~~q~)\\[4pt](~p~(q))\\[4pt]((p)~q~)\\[4pt]((p)(q))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}((p)(q))\\[4pt]((p)~q~)\\[4pt](~p~(q))\\[4pt](~p~~q~)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}((p)~q~)\\[4pt]((p)(q))\\[4pt](~p~~q~)\\[4pt](~p~(q))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(~p~(q))\\[4pt](~p~~q~)\\[4pt]((p)(q))\\[4pt]((p)~q~)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(~p~~q~)\\[4pt](~p~(q))\\[4pt]((p)~q~)\\[4pt]((p)(q))\end{matrix}}\!~}$ ${\displaystyle f_{15}\!~}$ ${\displaystyle ((~))\!~}$ ${\displaystyle ((~))\!~}$ ${\displaystyle ((~))\!~}$ ${\displaystyle ((~))\!~}$ ${\displaystyle ((~))\!~}$ ${\displaystyle {\text{Fixed Point Total}}\!~}$ ${\displaystyle 4\!~}$ ${\displaystyle 4\!~}$ ${\displaystyle 4\!~}$ ${\displaystyle 16\!~}$

 ${\displaystyle f\!~}$ ${\displaystyle \mathrm {D} f|_{\mathrm {d} p~\mathrm {d} q}\!~}$ ${\displaystyle \mathrm {D} f|_{\mathrm {d} p(\mathrm {d} q)}\!~}$ ${\displaystyle \mathrm {D} f|_{(\mathrm {d} p)\mathrm {d} q}\!~}$ ${\displaystyle \mathrm {D} f|_{(\mathrm {d} p)(\mathrm {d} q)}\!~}$ ${\displaystyle f_{0}\!~}$ ${\displaystyle (~)\!~}$ ${\displaystyle (~)\!~}$ ${\displaystyle (~)\!~}$ ${\displaystyle (~)\!~}$ ${\displaystyle (~)\!~}$ ${\displaystyle {\begin{matrix}f_{1}\\[4pt]f_{2}\\[4pt]f_{4}\\[4pt]f_{8}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(p)(q)\\[4pt](p)~q~\\[4pt]~p~(q)\\[4pt]~p~~q~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}((p,~q))\\[4pt]~(p,~q)~\\[4pt]~(p,~q)~\\[4pt]((p,~q))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(q)\\[4pt]~q~\\[4pt](q)\\[4pt]~q~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(p)\\[4pt](p)\\[4pt]~p~\\[4pt]~p~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(~)\\[4pt](~)\\[4pt](~)\\[4pt](~)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{3}\\[4pt]f_{12}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(p)\\[4pt]~p~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}((~))\\[4pt]((~))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}((~))\\[4pt]((~))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(~)\\[4pt](~)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(~)\\[4pt](~)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{6}\\[4pt]f_{9}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~(p,~q)~\\[4pt]((p,~q))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(~)\\[4pt](~)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}((~))\\[4pt]((~))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}((~))\\[4pt]((~))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(~)\\[4pt](~)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{5}\\[4pt]f_{10}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(q)\\[4pt]~q~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}((~))\\[4pt]((~))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(~)\\[4pt](~)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}((~))\\[4pt]((~))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(~)\\[4pt](~)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{7}\\[4pt]f_{11}\\[4pt]f_{13}\\[4pt]f_{14}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~(p~~q)~\\[4pt]~(p~(q))\\[4pt]((p)~q)~\\[4pt]((p)(q))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}((p,~q))\\[4pt]~(p,~q)~\\[4pt]~(p,~q)~\\[4pt]((p,~q))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~q~\\[4pt](q)\\[4pt]~q~\\[4pt](q)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~p~\\[4pt]~p~\\[4pt](p)\\[4pt](p)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(~)\\[4pt](~)\\[4pt](~)\\[4pt](~)\end{matrix}}\!~}$ ${\displaystyle f_{15}\!~}$ ${\displaystyle ((~))\!~}$ ${\displaystyle (~)\!~}$ ${\displaystyle (~)\!~}$ ${\displaystyle (~)\!~}$ ${\displaystyle (~)\!~}$

### Transforms Expanded over Ordinary Features

 ${\displaystyle f\!~}$ ${\displaystyle \mathrm {E} f|_{pq}\!~}$ ${\displaystyle \mathrm {E} f|_{p(q)}\!~}$ ${\displaystyle \mathrm {E} f|_{(p)q}\!~}$ ${\displaystyle \mathrm {E} f|_{(p)(q)}\!~}$ ${\displaystyle f_{0}\!~}$ ${\displaystyle (~)\!~}$ ${\displaystyle (~)\!~}$ ${\displaystyle (~)\!~}$ ${\displaystyle (~)\!~}$ ${\displaystyle (~)\!~}$ ${\displaystyle {\begin{matrix}f_{1}\\[4pt]f_{2}\\[4pt]f_{4}\\[4pt]f_{8}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(p)(q)\\[4pt](p)~q~\\[4pt]~p~(q)\\[4pt]~p~~q~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~\mathrm {d} p~~\mathrm {d} q~\\[4pt]~\mathrm {d} p~(\mathrm {d} q)\\[4pt](\mathrm {d} p)~\mathrm {d} q~\\[4pt](\mathrm {d} p)(\mathrm {d} q)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~\mathrm {d} p~(\mathrm {d} q)\\[4pt]~\mathrm {d} p~~\mathrm {d} q~\\[4pt](\mathrm {d} p)(\mathrm {d} q)\\[4pt](\mathrm {d} p)~\mathrm {d} q~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(\mathrm {d} p)~\mathrm {d} q~\\[4pt](\mathrm {d} p)(\mathrm {d} q)\\[4pt]~\mathrm {d} p~~\mathrm {d} q~\\[4pt]~\mathrm {d} p~(\mathrm {d} q)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(\mathrm {d} p)(\mathrm {d} q)\\[4pt](\mathrm {d} p)~\mathrm {d} q~\\[4pt]~\mathrm {d} p~(\mathrm {d} q)\\[4pt]~\mathrm {d} p~~\mathrm {d} q~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{3}\\[4pt]f_{12}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(p)\\[4pt]~p~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~\mathrm {d} p~\\[4pt](\mathrm {d} p)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~\mathrm {d} p~\\[4pt](\mathrm {d} p)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(\mathrm {d} p)\\[4pt]~\mathrm {d} p~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(\mathrm {d} p)\\[4pt]~\mathrm {d} p~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{6}\\[4pt]f_{9}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~(p,~q)~\\[4pt]((p,~q))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~(\mathrm {d} p,~\mathrm {d} q)~\\[4pt]((\mathrm {d} p,~\mathrm {d} q))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}((\mathrm {d} p,~\mathrm {d} q))\\[4pt]~(\mathrm {d} p,~\mathrm {d} q)~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}((\mathrm {d} p,~\mathrm {d} q))\\[4pt]~(\mathrm {d} p,~\mathrm {d} q)~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~(\mathrm {d} p,~\mathrm {d} q)~\\[4pt]((\mathrm {d} p,~\mathrm {d} q))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{5}\\[4pt]f_{10}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(q)\\[4pt]~q~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~\mathrm {d} q~\\[4pt](\mathrm {d} q)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(\mathrm {d} q)\\[4pt]~\mathrm {d} q~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~\mathrm {d} q~\\[4pt](\mathrm {d} q)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(\mathrm {d} q)\\[4pt]~\mathrm {d} q~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{7}\\[4pt]f_{11}\\[4pt]f_{13}\\[4pt]f_{14}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(~p~~q~)\\[4pt](~p~(q))\\[4pt]((p)~q~)\\[4pt]((p)(q))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}((\mathrm {d} p)(\mathrm {d} q))\\[4pt]((\mathrm {d} p)~\mathrm {d} q~)\\[4pt](~\mathrm {d} p~(\mathrm {d} q))\\[4pt](~\mathrm {d} p~~\mathrm {d} q~)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}((\mathrm {d} p)~\mathrm {d} q~)\\[4pt]((\mathrm {d} p)(\mathrm {d} q))\\[4pt](~\mathrm {d} p~~\mathrm {d} q~)\\[4pt](~\mathrm {d} p~(\mathrm {d} q))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(~\mathrm {d} p~(\mathrm {d} q))\\[4pt](~\mathrm {d} p~~\mathrm {d} q~)\\[4pt]((\mathrm {d} p)(\mathrm {d} q))\\[4pt]((\mathrm {d} p)~\mathrm {d} q~)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(~\mathrm {d} p~~\mathrm {d} q~)\\[4pt](~\mathrm {d} p~(\mathrm {d} q))\\[4pt]((\mathrm {d} p)~\mathrm {d} q~)\\[4pt]((\mathrm {d} p)(\mathrm {d} q))\end{matrix}}\!~}$ ${\displaystyle f_{15}\!~}$ ${\displaystyle ((~))\!~}$ ${\displaystyle ((~))\!~}$ ${\displaystyle ((~))\!~}$ ${\displaystyle ((~))\!~}$ ${\displaystyle ((~))\!~}$

 ${\displaystyle f\!~}$ ${\displaystyle \mathrm {D} f|_{pq}\!~}$ ${\displaystyle \mathrm {D} f|_{p(q)}\!~}$ ${\displaystyle \mathrm {D} f|_{(p)q}\!~}$ ${\displaystyle \mathrm {D} f|_{(p)(q)}\!~}$ ${\displaystyle f_{0}\!~}$ ${\displaystyle (~)\!~}$ ${\displaystyle (~)\!~}$ ${\displaystyle (~)\!~}$ ${\displaystyle (~)\!~}$ ${\displaystyle (~)\!~}$ ${\displaystyle {\begin{matrix}f_{1}\\[4pt]f_{2}\\[4pt]f_{4}\\[4pt]f_{8}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(p)(q)\\[4pt](p)~q~\\[4pt]~p~(q)\\[4pt]~p~~q~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~~\mathrm {d} p~~\mathrm {d} q~~\\[4pt]~~\mathrm {d} p~(\mathrm {d} q)~\\[4pt]~(\mathrm {d} p)~\mathrm {d} q~~\\[4pt]((\mathrm {d} p)(\mathrm {d} q))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~~\mathrm {d} p~(\mathrm {d} q)~\\[4pt]~~\mathrm {d} p~~\mathrm {d} q~~\\[4pt]((\mathrm {d} p)(\mathrm {d} q))\\[4pt]~(\mathrm {d} p)~\mathrm {d} q~~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~(\mathrm {d} p)~\mathrm {d} q~~\\[4pt]((\mathrm {d} p)(\mathrm {d} q))\\[4pt]~~\mathrm {d} p~~\mathrm {d} q~~\\[4pt]~~\mathrm {d} p~(\mathrm {d} q)~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}((\mathrm {d} p)(\mathrm {d} q))\\[4pt]~(\mathrm {d} p)~\mathrm {d} q~~\\[4pt]~~\mathrm {d} p~(\mathrm {d} q)~\\[4pt]~~\mathrm {d} p~~\mathrm {d} q~~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{3}\\[4pt]f_{12}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(p)\\[4pt]~p~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}\mathrm {d} p\\[4pt]\mathrm {d} p\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}\mathrm {d} p\\[4pt]\mathrm {d} p\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}\mathrm {d} p\\[4pt]\mathrm {d} p\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}\mathrm {d} p\\[4pt]\mathrm {d} p\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{6}\\[4pt]f_{9}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~(p,~q)~\\[4pt]((p,~q))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(\mathrm {d} p,~\mathrm {d} q)\\[4pt](\mathrm {d} p,~\mathrm {d} q)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(\mathrm {d} p,~\mathrm {d} q)\\[4pt](\mathrm {d} p,~\mathrm {d} q)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(\mathrm {d} p,~\mathrm {d} q)\\[4pt](\mathrm {d} p,~\mathrm {d} q)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(\mathrm {d} p,~\mathrm {d} q)\\[4pt](\mathrm {d} p,~\mathrm {d} q)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{5}\\[4pt]f_{10}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(q)\\[4pt]~q~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}\mathrm {d} q\\[4pt]\mathrm {d} q\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}\mathrm {d} q\\[4pt]\mathrm {d} q\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}\mathrm {d} q\\[4pt]\mathrm {d} q\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}\mathrm {d} q\\[4pt]\mathrm {d} q\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{7}\\[4pt]f_{11}\\[4pt]f_{13}\\[4pt]f_{14}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(~p~~q~)\\[4pt](~p~(q))\\[4pt]((p)~q~)\\[4pt]((p)(q))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}((\mathrm {d} p)(\mathrm {d} q))\\[4pt]~(\mathrm {d} p)~\mathrm {d} q~~\\[4pt]~~\mathrm {d} p~(\mathrm {d} q)~\\[4pt]~~\mathrm {d} p~~\mathrm {d} q~~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~(\mathrm {d} p)~\mathrm {d} q~~\\[4pt]((\mathrm {d} p)(\mathrm {d} q))\\[4pt]~~\mathrm {d} p~~\mathrm {d} q~~\\[4pt]~~\mathrm {d} p~(\mathrm {d} q)~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~~\mathrm {d} p~(\mathrm {d} q)~\\[4pt]~~\mathrm {d} p~~\mathrm {d} q~~\\[4pt]((\mathrm {d} p)(\mathrm {d} q))\\[4pt]~(\mathrm {d} p)~\mathrm {d} q~~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~~\mathrm {d} p~~\mathrm {d} q~~\\[4pt]~~\mathrm {d} p~(\mathrm {d} q)~\\[4pt]~(\mathrm {d} p)~\mathrm {d} q~~\\[4pt]((\mathrm {d} p)(\mathrm {d} q))\end{matrix}}\!~}$ ${\displaystyle f_{15}\!~}$ ${\displaystyle ((~))\!~}$ ${\displaystyle ((~))\!~}$ ${\displaystyle ((~))\!~}$ ${\displaystyle ((~))\!~}$ ${\displaystyle ((~))\!~}$

## Operational Representation

If you think that I linger in the realm of logical difference calculus out of sheer vacillation about getting down to the differential proper, it is probably out of a prior expectation that you derive from the art or the long-engrained practice of real analysis. But the fact is that ordinary calculus only rushes on to the sundry orders of approximation because the strain of comprehending the full import of ${\displaystyle \mathrm {E} \!~}$ and ${\displaystyle \mathrm {D} \!~}$ at once overwhelms its discrete and finite powers to grasp them. But here, in the fully serene idylls of zeroth order logic, we find ourselves fit with the compass of a wit that is all we'd ever need to explore their effects with care.

So let us do just that.

I will first rationalize the novel grouping of propositional forms in the last set of Tables, as that will extend a gentle invitation to the mathematical subject of group theory, and demonstrate its relevance to differential logic in a strikingly apt and useful way. The data for that account is contained in Table A3.

 ${\displaystyle f\!~}$ ${\displaystyle \mathrm {T} _{11}f\!~}$ ${\displaystyle \mathrm {E} f|_{\mathrm {d} p~\mathrm {d} q}\!~}$ ${\displaystyle \mathrm {T} _{10}f\!~}$ ${\displaystyle \mathrm {E} f|_{\mathrm {d} p(\mathrm {d} q)}\!~}$ ${\displaystyle \mathrm {T} _{01}f\!~}$ ${\displaystyle \mathrm {E} f|_{(\mathrm {d} p)\mathrm {d} q}\!~}$ ${\displaystyle \mathrm {T} _{00}f\!~}$ ${\displaystyle \mathrm {E} f|_{(\mathrm {d} p)(\mathrm {d} q)}\!~}$ ${\displaystyle f_{0}\!~}$ ${\displaystyle (~)\!~}$ ${\displaystyle (~)\!~}$ ${\displaystyle (~)\!~}$ ${\displaystyle (~)\!~}$ ${\displaystyle (~)\!~}$ ${\displaystyle {\begin{matrix}f_{1}\\[4pt]f_{2}\\[4pt]f_{4}\\[4pt]f_{8}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(p)(q)\\[4pt](p)~q~\\[4pt]~p~(q)\\[4pt]~p~~q~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~p~~q~\\[4pt]~p~(q)\\[4pt](p)~q~\\[4pt](p)(q)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~p~(q)\\[4pt]~p~~q~\\[4pt](p)(q)\\[4pt](p)~q~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(p)~q~\\[4pt](p)(q)\\[4pt]~p~~q~\\[4pt]~p~(q)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(p)(q)\\[4pt](p)~q~\\[4pt]~p~(q)\\[4pt]~p~~q~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{3}\\[4pt]f_{12}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(p)\\[4pt]~p~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~p~\\[4pt](p)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~p~\\[4pt](p)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(p)\\[4pt]~p~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(p)\\[4pt]~p~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{6}\\[4pt]f_{9}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~(p,~q)~\\[4pt]((p,~q))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~(p,~q)~\\[4pt]((p,~q))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}((p,~q))\\[4pt]~(p,~q)~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}((p,~q))\\[4pt]~(p,~q)~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~(p,~q)~\\[4pt]((p,~q))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{5}\\[4pt]f_{10}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(q)\\[4pt]~q~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~q~\\[4pt](q)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(q)\\[4pt]~q~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}~q~\\[4pt](q)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(q)\\[4pt]~q~\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}f_{7}\\[4pt]f_{11}\\[4pt]f_{13}\\[4pt]f_{14}\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(~p~~q~)\\[4pt](~p~(q))\\[4pt]((p)~q~)\\[4pt]((p)(q))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}((p)(q))\\[4pt]((p)~q~)\\[4pt](~p~(q))\\[4pt](~p~~q~)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}((p)~q~)\\[4pt]((p)(q))\\[4pt](~p~~q~)\\[4pt](~p~(q))\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(~p~(q))\\[4pt](~p~~q~)\\[4pt]((p)(q))\\[4pt]((p)~q~)\end{matrix}}\!~}$ ${\displaystyle {\begin{matrix}(~p~~q~)\\[4pt](~p~(q))\\[4pt]((p)~q~)\\[4pt]((p)(q))\end{matrix}}\!~}$ ${\displaystyle f_{15}\!~}$ ${\displaystyle ((~))\!~}$ ${\displaystyle ((~))\!~}$ ${\displaystyle ((~))\!~}$ ${\displaystyle ((~))\!~}$ ${\displaystyle ((~))\!~}$ ${\displaystyle {\text{Fixed Point Total}}\!~}$ ${\displaystyle 4\!~}$ ${\displaystyle 4\!~}$ ${\displaystyle 4\!~}$ ${\displaystyle 16\!~}$

The shift operator ${\displaystyle \mathrm {E} \!~}$ can be understood as enacting a substitution operation on the propositional form ${\displaystyle f(p,q)\!~}$ that is given as its argument. In our present focus on propositional forms that involve two variables, we have the following type specifications and definitions:

 ${\displaystyle {\begin{array}{lcl}\mathrm {E} ~:~(X\to \mathbb {B} )&\to &(\mathrm {E} X\to \mathbb {B} )\\[6pt]\mathrm {E} ~:~f(p,q)&\mapsto &\mathrm {E} f(p,q,\mathrm {d} p,\mathrm {d} q)\\[6pt]\mathrm {E} f(p,q,\mathrm {d} p,\mathrm {d} q)&=&f(p+\mathrm {d} p,q+\mathrm {d} q)\\[6pt]&=&f({\texttt {(}}p,\mathrm {d} p{\texttt {)}},{\texttt {(}}q,\mathrm {d} q{\texttt {)}})\end{array}}\!~}$

Evaluating ${\displaystyle \mathrm {E} f\!~}$ at particular values of ${\displaystyle \mathrm {d} p\!~}$ and ${\displaystyle \mathrm {d} q,\!~}$ for example, ${\displaystyle \mathrm {d} p=i\!~}$ and ${\displaystyle \mathrm {d} q=j,\!~}$ where ${\displaystyle i\!~}$ and ${\displaystyle j\!~}$ are values in ${\displaystyle \mathbb {B} ,\!~}$ produces the following result:

 ${\displaystyle {\begin{array}{lclcl}\mathrm {E} _{ij}&:&(X\to \mathbb {B} )&\to &(X\to \mathbb {B} )\\[6pt]\mathrm {E} _{ij}&:&f&\mapsto &\mathrm {E} _{ij}f\\[6pt]\mathrm {E} _{ij}f&=&\mathrm {E} f|_{\mathrm {d} p=i,\mathrm {d} q=j}&=&f(p+i,q+j)\\[6pt]&&&=&f({\texttt {(}}p,i{\texttt {)}},{\texttt {(}}q,j{\texttt {)}})\end{array}}\!~}$

The notation is a little awkward, but the data of Table A3 should make the sense clear. The important thing to observe is that ${\displaystyle \mathrm {E} _{ij}\!~}$ has the effect of transforming each proposition ${\displaystyle f:X\to \mathbb {B} \!~}$ into a proposition ${\displaystyle f^{\prime }:X\to \mathbb {B} .\!~}$ As it happens, the action of each ${\displaystyle \mathrm {E} _{ij}\!~}$ is one-to-one and onto, so the gang of four operators ${\displaystyle \{\mathrm {E} _{ij}:i,j\in \mathbb {B} \}\!~}$ is an example of what is called a transformation group on the set of sixteen propositions. Bowing to a longstanding local and linear tradition, I will therefore redub the four elements of this group as ${\displaystyle \mathrm {T} _{00},\mathrm {T} _{01},\mathrm {T} _{10},\mathrm {T} _{11},\!~}$ to bear in mind their transformative character, or nature, as the case may be. Abstractly viewed, this group of order four has the following operation table:

 ${\displaystyle \cdot \!~}$ ${\displaystyle \mathrm {T} _{00}\!~}$ ${\displaystyle \mathrm {T} _{01}\!~}$ ${\displaystyle \mathrm {T} _{10}\!~}$ ${\displaystyle \mathrm {T} _{11}\!~}$ ${\displaystyle \mathrm {T} _{00}\!~}$ ${\displaystyle \mathrm {T} _{00}\!~}$ ${\displaystyle \mathrm {T} _{01}\!~}$ ${\displaystyle \mathrm {T} _{10}\!~}$ ${\displaystyle \mathrm {T} _{11}\!~}$ ${\displaystyle \mathrm {T} _{01}\!~}$ ${\displaystyle \mathrm {T} _{01}\!~}$ ${\displaystyle \mathrm {T} _{00}\!~}$ ${\displaystyle \mathrm {T} _{11}\!~}$ ${\displaystyle \mathrm {T} _{10}\!~}$ ${\displaystyle \mathrm {T} _{10}\!~}$ ${\displaystyle \mathrm {T} _{10}\!~}$ ${\displaystyle \mathrm {T} _{11}\!~}$ ${\displaystyle \mathrm {T} _{00}\!~}$ ${\displaystyle \mathrm {T} _{01}\!~}$ ${\displaystyle \mathrm {T} _{11}\!~}$ ${\displaystyle \mathrm {T} _{11}\!~}$ ${\displaystyle \mathrm {T} _{10}\!~}$ ${\displaystyle \mathrm {T} _{01}\!~}$ ${\displaystyle \mathrm {T} _{00}\!~}$

It happens that there are just two possible groups of 4 elements. One is the cyclic group ${\displaystyle Z_{4}\!~}$