[MD] Idealistic static value patterns

[mail at tuukkavirtaperko.net]

Here's a legible version of the previous:

T = {Tm | 1 ≤ m ≤ n}

This splits each set T into n patterns. n = 4 is the traditional Pirsig
solution.

     Nm ⊆ Sm
     Nm ⊆ Om

Here normative patterns are included in the descriptive patterns. They
are supposed to explain the structure of the descriptive patterns.

℘(Rm) = Sn – (m – 1) ∩ Om

This means that subjective and objective quality emerge from romantic
quality in addition to emerging from each other. And like the picture
(http://moq.fi/CM-2.png) shows, subsets of subjective quality S emerge
from romantic quality R in inverse order compared to subsets of
objective quality O. This explains why n = 1 would comprise a dull
theory. The inversion of order could not be noticed if both S and O had
only one pattern!

The power set function ℘ means emergence. The power set of any set T
includes all subsets of that set, but not necessarily the elements of
the subsets. Say you have a set T with elements:

a,b,c

The power set of that set T would contain the following sets:

T0 = {} (empty set)
T1 = {c}
T2 = {b}
T3 = {b,c}
T4 = {a}
T5 = {a,c}
T6 = {a,b}
T7 = {a,b,c}

Power sets do not necessarily contain the elements of the subsets, only
the subsets. This explains why normative quality does not contain
sensory perceptions despite emerging from intellectual quality, which
does.

Let's examine the following snippet:

BEGIN PASTE

     1I :=def O
     2I :=def N
     3I :=def S

Let ℘(T) denote the powerset of an arbitrary set T in our set theory.
Let us define a number-theoretic function Modn:ℕ → ℕ ∩ [0,n[ of one  
free variable such that for every p ∈ ℤ+ and q ∈ ℕ:

    Modp(q) = q – p⌊q⁄p⌋

Thus, Modp(q) is the remainder that results when a natural number q is
divided by a positive integer p. Let the following two formulas of our
set theory hold:

     ∀k(k ∈ {1,2,3} ⇒ ∀m(m ∈ ℕ ∩ [1,n[ ⇒ ℘(kIm) ⊆ kIm + 1))
     ∀k(k ∈ {1,2,3} ⇒ ℘(kIn) ⊆ 1 + Mod3(k)I1))



END PASTE

This basically formalizes the notion that each pattern emerges from the
pattern preceding it. The 1I, 2I and 3I simply give index names to
objective quality, normative quality and subjective quality. The cycle
uses these index names to refer to them.

This line:

     ∀k(k ∈ {1,2,3} ⇒ ∀m(m ∈ ℕ ∩ [1,n[ ⇒ ℘(kIm) ⊆ kIm + 1))

makes the cycle run within one pattern system, such as objective
quality. But it does not make the cycle shift from one pattern system
to another. That is done by this line:

     ∀k(k ∈ {1,2,3} ⇒ ℘(kIn) ⊆ 1 + Mod3(k)I1))

Mod3kI divides the index number of I with 3, gets the remainder and
adds one. In effect:

1I => remainder of 1/3, which is 1 => add 1 to that => 2
2I => remainder of 2/3, which is 2 => add 1 to that => 3
3I => remainder of 3/3, which is 0 => add 1 to that => 1

By the way, I uncovered two mistakes in the formulas while explaining
this to you, so this is certainly a very useful activity! They weren't
due to Timo but due to my own mistakes.

What does this do then?

    Modp(q) = q – p⌊q⁄p⌋

It defines a function. This function performs the calculation q/p and
applies the floor function to the result. The floor function is denoted
by ⌊ and ⌋ and it practically just rounds down the input it receives.

1/3 = 0,3333... => apply floor function => 0
2/3 = 0,6666... => apply floor function => 0
3/3 = 1 => apply floor function => 1

This sort of works as a trigger. Whenever the floor function outputs 0,
p is not "activated" because p * 0 = 0. But if we are at the last
indexed pattern 3I, the floor function outputs 1. As we all know, p * 1
= p.

p could stand for "payload", if you will. (BOOM!!)

Whenever payload is not activated, Modp is an innocuous function. You
input a number, and it outputs the same number. But when payload is
activated at the end of the cycle, it outputs 0.

Basically, the Modp function doesn't do more than that. But see this
formula again:

     ∀k(k ∈ {1,2,3} ⇒ ℘(kIn) ⊆ 1 + Mod3(k)I1))

In this formula, 1 is added to anything Modp outputs. So if Modp
outputs the same number, this function nevertheless adds one to it. But
if Modp outputs 0, this function again adds 1, which is necessary,
because 0I does not correspond to any defined form of quality.

Enough of the formalisms.