ONT Differential Logic -- Series A
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DLOG. Note A1
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One of the first things that you can do, once you
have a really decent calculus for boolean functions
or propositional logic, whatever you want to call it,
is to compute the differentials of these functions or
propositions.
Now there are many ways to dance around this idea,
and I feel like I have tried them all, before one
gets down to acting on it, and there many issues
of interpretation and justification that we will
have to clear up after the fact, that is, before
we can be sure that it all really makes any sense,
but I think this time I'll just jump in, and show
you the form in which this idea first came to me.
Start with a proposition of the form x & y, which
I graph as two labels attached to a root node, so:
o-------------------------------------------------o
| |
| x y |
| @ |
| |
o-------------------------------------------------o
| x and y |
o-------------------------------------------------o
Written as a string, this is just the concatenation "x y".
The proposition xy may be taken as a boolean function f(x, y)
having the abstract type f : B x B -> B, where B = {0, 1} is
read in such a way that 0 means "false" and 1 means "true".
In this style of graphical representation,
the value "true" looks like a blank label
and the value "false" looks like an edge.
o-------------------------------------------------o
| |
| |
| @ |
| |
o-------------------------------------------------o
| true |
o-------------------------------------------------o
o-------------------------------------------------o
| |
| o |
| | |
| @ |
| |
o-------------------------------------------------o
| false |
o-------------------------------------------------o
Back to the proposition xy. Imagine yourself standing
in a fixed cell of the corresponding venn diagram, say,
the cell where the proposition xy is true, as pictured:
o-------------------------------------------------o
| |
| |
| o-----------o o-----------o |
| / \ / \ |
| / o \ |
| / /%\ \ |
| / /%%%\ \ |
| o o%%%%%o o |
| | |%%%%%| | |
| | |%%%%%| | |
| | x |%%%%%| y | |
| | |%%%%%| | |
| | |%%%%%| | |
| o o%%%%%o o |
| \ \%%%/ / |
| \ \%/ / |
| \ o / |
| \ / \ / |
| o-----------o o-----------o |
| |
| |
o-------------------------------------------------o
Now ask yourself: What is the value of the
proposition xy at a distance of dx and dy
from the cell xy where you are standing?
Don't think about it -- just compute:
o-------------------------------------------------o
| |
| dx o o dy |
| / \ / \ |
| x o---@---o y |
| |
o-------------------------------------------------o
| (x + dx) and (y + dy) |
o-------------------------------------------------o
To make future graphs easier to draw in Ascii land,
I will use devices like @=@=@ and o=o=o to identify
several nodes into one, as in this next redrawing:
o-------------------------------------------------o
| |
| x dx y dy |
| o---o o---o |
| \ | | / |
| \ | | / |
| \| |/ |
| @=@ |
| |
o-------------------------------------------------o
| (x + dx) and (y + dy) |
o-------------------------------------------------o
However you draw it, these expressions follow because the
expression x + dx, where the plus sign indicates (mod 2)
addition in B, and thus corresponds to an exclusive-or
in logic, parses to a graph of the following form:
o-------------------------------------------------o
| |
| x dx |
| o---o |
| \ / |
| @ |
| |
o-------------------------------------------------o
| x + dx |
o-------------------------------------------------o
Next question: What is the difference between
the value of the proposition xy "over there" and
the value of the proposition xy where you are, all
expressed as general formula, of course? Here 'tis:
o-------------------------------------------------o
| |
| x dx y dy |
| o---o o---o |
| \ | | / |
| \ | | / |
| \| |/ x y |
| o=o-----------o |
| \ / |
| \ / |
| \ / |
| \ / |
| \ / |
| \ / |
| @ |
| |
o-------------------------------------------------o
| ((x + dx) & (y + dy)) - xy |
o-------------------------------------------------o
Oh, I forgot to mention: Computed over B,
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
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:
o-------------------------------------------------o
| |
| dx dy |
| o---o o---o |
| \ | | / |
| \ | | / |
| \| |/ |
| o=o-----------o |
| \ / |
| \ / |
| \ / |
| \ / |
| \ / |
| \ / |
| @ |
| |
o-------------------------------------------------o
| ((1 + dx) & (1 + dy)) - 1&1 |
o-------------------------------------------------o
And this is equivalent to the following graph:
o-------------------------------------------------o
| |
| dx dy |
| o o |
| \ / |
| o |
| | |
| @ |
| |
o-------------------------------------------------o
| dx or dy |
o-------------------------------------------------o
Enough for the moment.
Explanation to follow.
Jon Awbrey
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http://www.cs.bsu.edu/homepages/mighty/history.html
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