Re: Mystical Validation Tested

Daniel Fabulich (daniel.fabulich@yale.edu)
Wed, 27 May 1998 23:41:09 -0400 (EDT)


On Wed, 27 May 1998, Ian Goddard wrote:

> Since there is never an instance of atomism, we cannot
> exactly say H = A only in that H always is and A never
> is. The process of inquiry must discover what is and
> what is not, and I posit that atomism always is not.
>
> Here "atomism" refers to the idea that identity is
> atomized such that the identity of A comes only from
> A and not from -A. I posit that such is never true.

I think we disagree as to what atomism signifies. Atomism is making a
statement about the actual values on the identity diagram, rather than
about their sum. For example, on your site you give two objects, A and B,
where B is one greater than A. If I add 0 to this mix, we get the
following identity matrix:

0 A B
0 0 +1 +2
A -1 0 +1
B -2 -1 0

When atomism states that A can change independently of B, it means that it
changes relative to 0, and that this does not affect B relative to 0. For
example, if A increases by three relative to 0, we get the new chart:

0 A B
0 0 +4 +2
A -4 0 -2
B -2 +2 0

That this change is possible is trivially true; that it is compatible with
atomism is also true. Not even atomism claims that something can change
relative to itself, but rather that it changes relative to 0. Note that
this does not affect B relative to 0 (+2), or 0 relative to B (-2), which
was all that atomism was trying to prove anyway.

> IAN: If identity is difference, then A = 0 and -A = 0,
> where each are differentiated from themselves, which
> is to say that the difference between A and A = 0.
>
> If identity is difference, then A is A only by differ-
> entaiton from -A. Lo and behold, the sum of that rela-
> tion also = 0. We cannot segment that relation into
> two separate parts without each becoming 0, and
> yet together they are also 0... in sum.

Wait. Stop there. What I'm trying to say is that under the mystical
experience, the difference between 0 and 1 is zero, NOT that the sum of
the difference between 0 and 1 and the difference between 1 and 0 is zero.
The latter is trivial; the former is mystical.

A mysticist's identity chart would look like this:

0 1
0 0 0
1 0 0

or, since 0 = 1,

0 1
0 1 1
1 1 1

or even

0 1
0 1 1
1 0 0

That your other identity charts do NOT look like this indicates strongly
to me that the difference between 1 and 0 is 1, and that the difference
between 0 and 1 is -1. Though these SUM to 0, my mystical identity charts
differ from your identity charts in a non-trivial way. I assert that MY
identity chart is mystical, whereas yours is not. :)

> The showing of "all such values actually being
> zero" is the fact that "self to self = 0," that
> is, that the difference/identity of A to A = 0,
> which holds for every thing. Each thing gets a
> non-zero identity only by giving the opposite
> non-zero to another thing (or empty space) and
> the sum of that relation is the same as the self
> to self relation. So there is never not-zero.

I should have been more specific; I mean that you must show that 1 is 0
relative to 0 in order to demonstrate mysticism, as per the identity
charts I drew earlier; anything less is simply the regular old physics we
know and love.

> Zero may also be defined as "symmetry."
> If there is asymmetry, it is what it is
> relative to symmetry, and that's symmetry!
> So there can never be not-symmetry in all.

Er, *I* don't define 0 as symmetry, and I don't think it's valid to do so.
Sure, symmetrical things will sum to zero in the way you describe, but you
commit a fallacy of parts to say that just because two numbers sum to a
value that both are that value. This is like saying that since 1 + 2 = 3,
1 = 3 and 2 = 3. Clearly this is not true; but in the same way, if 1 + -1
= 0, you cannot say that 1 = 0.

> IAN: I think that your saying that if holism always
> is, such that there is no not-holism, this is atomism,
> i.e., there is H and H is H free from not-H.

On the contrary, I am simply saying that H = A, a conclusion which you
disagreed with earlier but which I assert.