Cf: Differential Logic : 5
At:
http://inquiryintoinquiry.com/2020/03/28/differential-logic-%e2%80%a2-5/
Note. I gave it the old college try at transcribing
the following math formulas but I recommend following
the link above for a much more readable copy.
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.
Venn Diagram f = pq
https://inquiryintoinquiry.files.wordpress.com/2020/03/venn-diagram-f-p-and-q.jpg
Cactus Graph f = pq
https://inquiryintoinquiry.files.wordpress.com/2020/03/cactus-graph-f-p-and-q.jpg
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 : B x B -> B or f : B^2 -> B. The concrete type takes into account the qualitative dimensions or the "units" of
the case, which can be explained as follows.
* Let P be the set of values {(p), p} = {not p, p} isomorphic to B = {0, 1}.
* Let Q be the set of values {(q), q} = {not q, q} isomorphic to B = {0, 1}.
Then interpret the usual propositions about p, q as functions of the concrete type f : P x Q -> B.
We are going to consider various operators on these functions. An operator F is a function which takes one function f
into another function Ff.
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 D.
* The enlargement operator Epsilon, written here as E.
These days, 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 x Q, its "(first order) differential extension" EX is constructed
according to the following specifications:
* EX = X x dX
where:
* X = P x Q
* dX = dP x dQ
* dP = {(dp), dp}
* dQ = {(dq), dq}
The interpretations of these new symbols can be diverse, but the easiest option for now is just to say dp means "change
p" and dq means "change q".
Drawing a venn diagram for the differential extension EX = X x dX requires four logical dimensions, P, Q, dP, dQ, but it
is possible to project a suggestion of what the differential features dp and dq are about on the 2-dimensional base
space X = P x Q by drawing arrows that cross the boundaries of the basic circles in the venn diagram for X, reading an
arrow as dp if it crosses the boundary between p and (p) in either direction and reading an arrow as dq if it crosses
the boundary between q and (q) in either direction, as indicated in the following figure.
Venn Diagram p q dp dq
https://inquiryintoinquiry.files.wordpress.com/2020/03/venn-diagram-p-q-dp-dq.jpg
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 (dp (dq)) says the same thing as dp => dq, 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 x Q, the "(first order) enlargement of f" is the proposition Ef over
the differential extension EX defined by the following formula:
* Ef(p, q, dp, dq)
= f(p + dp, q + dq)
= f(p xor dp, q xor dq)
In the example f(p, q) = pq, the enlargement Ef is computed as follows:
* Ef(p, q, dp, dq)
= (p + dp)(q + dq)
= (p xor dp)(q xor dq)
The corresponding cactus graph is shown below.
Cactus Graph Ef = (p,dp)(q,dq)
https://inquiryintoinquiry.files.wordpress.com/2020/03/cactus-graph-ef-pdpqdq.jpg
Given the proposition f(p, q) over X = P x Q, the "(first order) difference of f" is the proposition Df over EX defined
by the formula Df = Ef - f, or, written out in full:
* Df(p, q, dp, dq)
= f(p + dp, q + dq) - f(p, q)
= f(p xor dp, q xor dq) xor f(p, q)
In the example f(p, q) = pq, the difference Df is computed as follows:
* Df(p, q, dp, dq)
= (p + dp)(q + dq) - pq
= (p xor dp)(q xor dq) xor pq
The corresponding cactus graph is shown below.
Cactus Graph Df = ((p,dp)(q,dq),pq)
https://inquiryintoinquiry.files.wordpress.com/2020/03/cactus-graph-df-pdpqdqpq.jpg
At the end of the previous section we evaluated this first order difference of conjunction Df 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 Df|_{pq} or Df|_{(1, 1)}, and we arrived at the
locally applicable law which may be stated and illustrated as follows:
* f(p, q) = pq = p and q => Df|_{pq} = ((dp)(dq)) = dp or dq
Venn Diagram Difference pq @ pq
https://inquiryintoinquiry.files.wordpress.com/2020/03/venn-diagram-pq-difference-pq-40-pq-1.jpg
Cactus Graph Difference pq @ pq
https://inquiryintoinquiry.files.wordpress.com/2020/03/cactus-graph-pq-difference-pq-40-pq.jpg
The venn diagram shows the analysis of the inclusive disjunction "dp or dq" into the following exclusive disjunction:
* (dp and not dq) xor (dq and not dp) xor (dp and dq)
The resultant differential proposition may be read to say "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.
Regards,
Jon