12,080
edits
Changes
MyWikiBiz, Author Your Legacy — Wednesday May 01, 2024
Jump to navigationJump to searchDirectory:Jon Awbrey/Papers/Cactus Language (view source)
Revision as of 18:23, 27 May 2009
, 18:23, 27 May 2009→Differential Logic: ASCII → JPEG
Start with a proposition of the form <math>x ~\operatorname{and}~ y,</math> which is graphed as two labels attached to a root node:
Start with a proposition of the form <math>x ~\operatorname{and}~ y,</math> which is graphed as two labels attached to a root node:
{| align="center" cellpadding="6" width="90%"
{| align="center" cellpadding="6" style="text-align:center" width="90%"
|
|
<pre>
<pre>
In this style of graphical representation, the value <math>\operatorname{true}</math> looks like a blank label and the value <math>\operatorname{false}</math> looks like an edge.
In this style of graphical representation, the value <math>\operatorname{true}</math> looks like a blank label and the value <math>\operatorname{false}</math> looks like an edge.
{| align="center" cellpadding="6" width="90%"
{| align="center" cellpadding="6" style="text-align:center" width="90%"
|
|
<pre>
<pre>
|}
|}
{| align="center" cellpadding="6" width="90%"
{| align="center" cellpadding="6" style="text-align:center" width="90%"
|
|
<pre>
<pre>
in a fixed cell of the corresponding venn diagram, say, the cell where the proposition <math>xy\!</math> is true, as shown here:
in a fixed cell of the corresponding venn diagram, say, the cell where the proposition <math>xy\!</math> is true, as shown here:
{| align="center" cellpadding="6" width="90%"
{| align="center" cellpadding="10" style="text-align:center"
|
| [[Image:Venn_Diagram_X_And_Y.jpg|500px]]
|}
|}
Don't think about it — just compute:
Don't think about it — just compute:
{| align="center" cellpadding="6" width="90%"
{| align="center" cellpadding="6" style="text-align:center" width="90%"
|
|
<pre>
<pre>
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:
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:
{| align="center" cellpadding="6" width="90%"
{| align="center" cellpadding="6" style="text-align:center" width="90%"
|
|
<pre>
<pre>
However you draw it, these expressions follow because the expression <math>x + dx,\!</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:
However you draw it, these expressions follow because the expression <math>x + dx,\!</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:
{| align="center" cellpadding="6" width="90%"
{| align="center" cellpadding="6" style="text-align:center" width="90%"
|
|
<pre>
<pre>
|}
|}
Next question: What is the difference between the value of the proposition <math>xy\!</math> "over there" and the value of the proposition <math>xy\!</math> where you are, all expressed as general formula, of course? Here 'tis:
{| align="center" cellpadding="6" style="text-align:center" width="90%"
|
<pre>
<pre>
o---------------------------------------o
o---------------------------------------o
| |
| |
| ((x + dx) & (y + dy)) - xy |
| ((x + dx) & (y + dy)) - xy |
o---------------------------------------o
o---------------------------------------o
</pre>
|}
Oh, I forgot to mention: Computed over B,
Oh, I forgot to mention: Computed over <math>\mathbb{B},</math> plus and minus are the very same operation. This will make the relationship between the differential and the integral parts of the resulting calculus slightly stranger than usual, but never mind that now.
plus and minus are the very same operation.
This will make the relationship between the
differential and the integral parts of the
resulting calculus slightly stranger than
usual, but never mind that now.
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>xy\!</math> is true? Well, substituting <math>1\!</math> for <math>x\!</math> and <math>1\!</math> for <math>y\!</math> in the graph amounts to the same thing as erasing those labels:
from your current standpoint, that is, evaluated at the point
where xy is true? Well, substituting 1 for x and 1 for y in
the graph amounts to the same thing as erasing those labels:
{| align="center" cellpadding="6" style="text-align:center" width="90%"
|
<pre>
o---------------------------------------o
o---------------------------------------o
| |
| |
| ((1 + dx) & (1 + dy)) - 1·1 |
| ((1 + dx) & (1 + dy)) - 1·1 |
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" style="text-align:center" width="90%"
|
<pre>
o---------------------------------------o
o---------------------------------------o
| |
| |
| dx or dy |
| dx or dy |
o---------------------------------------o
o---------------------------------------o
</pre>
|}
Have to break here -- will explain later.
Have to break here — will explain later.
====Note 2====
====Note 2====
