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I am going to excerpt some of my previous explorations on differential logic and dynamic systems and bring them to bear on the sorts of discrete dynamical themes that we find of interest in the NKS Forum. This adaptation draws on the "Cactus Rules", "Propositional Equation Reasoning Systems", and "Reductions Among Relations" threads, and will in time be applied to the "Differential Analytic Turing Automata" thread:
I am going to excerpt some of my previous explorations on differential logic and dynamic systems and bring them to bear on the sorts of discrete dynamical themes that we find of interest in the NKS Forum. This adaptation draws on the "Cactus Rules", "Propositional Equation Reasoning Systems", and "Reductions Among Relations" threads, and will in time be applied to the "Differential Analytic Turing Automata" thread:
* CR. http://forum.wolframscience.com/showthread.php?threadid=256
:* CR. http://forum.wolframscience.com/showthread.php?threadid=256
* PERS. http://forum.wolframscience.com/showthread.php?threadid=297
:* PERS. http://forum.wolframscience.com/showthread.php?threadid=297
* RAR. http://forum.wolframscience.com/showthread.php?threadid=400
:* RAR. http://forum.wolframscience.com/showthread.php?threadid=400
* DATA. http://forum.wolframscience.com/showthread.php?threadid=228
:* DATA. http://forum.wolframscience.com/showthread.php?threadid=228
One of the first things that you can do, once you have a moderately functional calculus for boolean functions or propositional logic, whatever you choose to call it, is to start thinking about, and even start computing, the differentials of these functions or propositions.
One of the first things that you can do, once you have a moderately functional calculus for boolean functions or propositional logic, whatever you choose to call it, is to start thinking about, and even start computing, the differentials of these functions or propositions.
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<pre>
Now ask yourself: What is the value of the proposition <math>pq\!</math> at a distance of <math>\operatorname{d}p</math> and <math>\operatorname{d}q</math> from the cell <math>pq\!</math> where you are standing?
Don't think about it — just compute:
Don't think about it -- just compute:
{| align="center" cellpadding="6" width="90%"
| align="center" |
<pre>
o-------------------------------------------------o
o-------------------------------------------------o
| |
| |
| (p, dp) (q, dq) |
| (p, dp) (q, dq) |
o-------------------------------------------------o
o-------------------------------------------------o
</pre>
|}
To make future graphs easier to draw in Asciiland,
To make future graphs easier to draw in ASCII, I will use devices like '''<code>@=@=@</code>''' and '''<code>o=o=o</code>''' to identify several nodes into one, as in this next redrawing:
I'll use devices like @=@=@ and o=o=o to identify
several nodes into one, as in this next redrawing:
{| align="center" cellpadding="6" width="90%"
| align="center" |
<pre>
o-------------------------------------------------o
o-------------------------------------------------o
| |
| |
| (p, dp) (q, dq) |
| (p, dp) (q, dq) |
o-------------------------------------------------o
o-------------------------------------------------o
</pre>
|}
However you draw it, these expressions follow because the
However you draw it, these expressions follow because the expression <math>p + \operatorname{d}p,</math> where the plus sign indicates addition in <math>\mathbb{B},</math> that is, addition modulo 2, and thus corresponds to the exclusive disjunction operation in logic, parses to a graph of the following form:
expression p + dp, where the plus sign indicates addition
in B and thus corresponds to the exclusive-or in logic,
parses to a graph of the following form:
{| align="center" cellpadding="6" width="90%"
| align="center" |
<pre>
o-------------------------------------------------o
o-------------------------------------------------o
| |
| |
| (p, dp) |
| (p, dp) |
o-------------------------------------------------o
o-------------------------------------------------o
</pre>
|}
Next question: What is the difference between the value of the
Next question: What is the difference between the value of the
proposition pq "over there", at a remove of dp dq, and the value
proposition <math>pq\!</math> "over there", at a remove of <math>\operatorname{d}p~\operatorname{d}q,\!</math> and the value of the proposition <math>pq\!</math> where you are, all expressed in the form of
of the proposition pq where you are, all expressed in the form of
a general formula, of course? Here is the appropriate formulation:
a general formula, of course? Here is the appropriate formulation:
{| align="center" cellpadding="6" width="90%"
| align="center" |
<pre>
o-------------------------------------------------o
o-------------------------------------------------o
| |
| |
| ((p, dp)(q, dq), p q) |
| ((p, dp)(q, dq), p q) |
o-------------------------------------------------o
o-------------------------------------------------o
</pre>
|}
There is one thing that I ought to mention at this point:
There is one thing that I ought to mention at this point: Computed over <math>\mathbb{B},</math> 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.
Computed over 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
Last question, for now: What is the value of this expression from your current standpoint, that is, evaluated at the point where <math>pq\!</math> is true? Well, substituting <math>1\!</math> for <math>p\!</math> and <math>1\!</math> for <math>q\!</math> in the graph amounts to erasing the labels <math>p\!</math> and <math>q\!,</math> as shown here:
from your current standpoint, that is, evaluated at the point
where pq is true? Well, substituting 1 for p and 1 for q in
the graph amounts to erasing the labels "p" and "q", like so:
{| align="center" cellpadding="6" width="90%"
| align="center" |
<pre>
o-------------------------------------------------o
o-------------------------------------------------o
| |
| |
| (( , dp)( , dq), ) |
| (( , dp)( , dq), ) |
o-------------------------------------------------o
o-------------------------------------------------o
</pre>
|}
And this is equivalent to the following graph:
And this is equivalent to the following graph:
{| align="center" cellpadding="6" width="90%"
| align="center" |
<pre>
o-------------------------------------------------o
o-------------------------------------------------o
| |
| |
o-------------------------------------------------o
o-------------------------------------------------o
</pre>
</pre>
|}
==Note 2==
==Note 2==