Metaphysics

Revisiting The Construction
Chapter two of the book introduced the sets u and v as a partition of the sets A and B into distinct sets of predicates / properties that were positively exemplified and positively nonexemplified. "A", consequent of q, became as either "r" or "u^{1}", "B" became from p&q^{1} as s or v^{1}.
This poses quite a problem, as x and x^{1} are not strictly functions on the "same" single predicate, but are treated as if they were conjunctions made upon totally disjoint sets. Logically, r&s entails both r and s and cannot be considered "disjoint" whereas (r&s)^{1} certainly may be! Also, the statement that putting u=v^{1} and v=u^{1} makes no difference to the structure of (or the local closure of) the K4 group in {p, u, v} (and then also {p, u^{1}, v^{1}}) is worthy of note: for the disjunction (p&u^{1}) v (p^{1}&v^{1}) rearranges to (p^{1}&u) v (p&v) inverting both sides.
So, r, u, s, v are all "disjoint" sets in the octal, yet only found in the octal in the form r, s, u^{1} and v^{1} (also (r&s)^{1} and u&v).
Now, returning to the construction: on separating A and B above, it is just as legal (given x and x^{1} are disjoint sets) to put A = {r : q=> r} from the disjunctrion q v p&q^{1}. Then , B = {s : p&q^{1}=>s}. This was how I began. However, then it is legal to split A and B once again in the form r = r1, s= s1, u=r1^{1}, v=s1^{1}. Then r&u^{1} = r1 and v&s^{1}=s1. Just as legally then I have the result of r1=s1^{1} and s1=r1^{1} (given Pos(p)).
Now, in the octal; were I not to have this last condition that r1=s1^{1}, I would find the disjunction (r&s)^{1} v (p&r&s)=>(u&v) to be inconsistent as u&v = (r&s)^{1}. Then r&s is equal to it's own inversion, a contradiction. In the proper action of the octal, the operation of virtue in this last disjunction fails completely if r=s^{1}. (This disjunction is of a different "local closure" that that of {p, r , s}.) That is, if u and v are disjoint sets from r and s, then putting r=s^{1} causes (r&s)^{1} v u&v to become inconsistent. In the case when u=r1^{1} etc, the octal initially appears inconsistent without that substitution.
Yet, it must be a certaintly that in the case of u=r1^{1} etc, the local closure is still the union of {p, r, s} and no new information is obtained from u and v. That the sets u and v are "extended" over and above, and from r and s inverted, is a simple "proof" that as (r&s)^{1} v u&v is equal both sides, (r&s)^{1} is inconsistent with the closure of the set {p, r, s} and is "empty" or rather, "not in the filter" or the indexing set.
So, a defining condition of the octal must be that if "r" contains "x" then "u" must never contain x^{1}. Or else there is no octal, and only the single K4 form. This seems unnerving but is due to the axiom of positive properties where I stated that if x is not able to be positively exemplified, then x^{1} is certainly positively exemplified.
So, the sets in the octal are {p, r, s, u, v, r&s, u&v} which properly partition Ω and these seven are all "disjoint": The K4 form of {p, r, s} is then closed, and the ultrafilter indexes every set in the octal to action or inaction to the schema of {p, r, s, u^{1}, v^{1}, (r&s)^{1}, u&v}.
I had stated that the positive properties themselves do not form an octal group: but as every disjunction x v (p&x^{1}=>y) may be inverted to y^{1} v (p&y=>x^{1}), this shows that every freely decidable disjunction is able to be inverted without privating virtue: and the octal itself whilst constant in positive properties, is indexed in a fashion where any positive property (except virtue) may be inverted and the schema will compensate for that!
I.e. if I invert r and s, I gain {p, r^{1}, s^{1}, r&s, u, v, (u&v)^{1}} and inverting in this manner shows that the octal indexed after this manner shows there is room for some "judgement".
There is one caveat then: that it is not strictly correct to state that x and x^{1} are separate predicates: as they are indeed applied to the same positive property. That said, the sets in {p, r, s, u^{1}, v^{1}, (r&s)^{1}, u&v} are still disjoint in the octal and also span the octal: For any local closure may be locally inverted and the sets within the octal {p, r, s, u, v, r&s, u&v} reobtained, in one form or another!
So {p, u^{1}, v^{1}} is actually a simple rearrangement of {p, u, v} and there is no contradiction in the structure of the octal, the division of the ultrafilter upon inclusion or exclusion (exemplifying x or x^{1} respectfully) is enough to guarantee the set of all positive properties is consistent. (And that the octal is "all of Ω".)
So, back once more to the construction:
If then, I put r=r1 and u=r1^{1} etc, I show there is no (consistent) middle (r&s)^{1}.
If I also put u^{1}=v and r^{1}=s etc, then this quandary disappears, only if there is no consistent middle (r&s)^{1}.
If I also place u^{1} = s^{1} and v^{1}=r^{1} which is equal to the above, every problem disappears, and there is no apparent need for virtue to decide the disjunction:
Then the K4 form is closed, and there are no "local" positive properties apart from {p, r, s}.
Then the answer is simply that if there is no octal posited over the K4 form, and the only separation is into r = u^{1} = r1, s = v^{1} = s1 etc, Then I will not find an octal, no matter how much I twist it! (I instead require virtue to also act on (r&s)^{1}. The octal exists by axiom of virtue only if it is found positive to perform r&s despite the empty middle in r v p&r^{1}=>s, then there must exist that virtue in p to entail some u&v. A lost opportunity sometimes occurs!)
Similarly, given there is always an octal over the K4 form, there will always be found a suitable "u" and a "v". (Such that "r" is never a predicate equal to u or u^{1}, and the sets in all positive properties of r and u^{1} as well as u and r^{1} are disjoint. One of each pair is included by the ultrafilter and the other is not.)
Then I have the condition that an octal exists only if I also have r and u, or s and v are not equals or inverses: (That is, up to rearranging and negating both sides in a disjunction of virtue in the set A v p&B as in the construction.)
Crucially I did not assume u and v "empty", merely that "A" is a set of positive properties as is "B", as is also the set X in all virtue p. And I have quite easily found that not only is this local closure truly closed (There is no deciding the conjunction (r&s)^{1} v (r&s)^{1}) but also I found that the existence of the octal "The Father" is constructed as a closure over and upon the K4 form, yet is not entailed from the K4 form in the special case where u and v are s and r respectfully. Although the K4 form as a subset of the Father's octal is logically entailed. (One is Father, the other His "Son", logically and structurally.)
The question then arises: If it is always seen that r&s is a positive property (in Anselm's filter), then r&s cannot private virtue. Then, is p&r&s => u&v always nonempty? If u&v has content not in (r&s)^{1} or r&s, and entails no virtue (is not virtue itself) then there exists an octal. This excludes the cases of u=r1^{1} etc from before.
Yet even then, only (r&s)^{1} is positively exemplified in the octal. (r&s)^{1} is never found in any closed K4 form with p and one of r and s.
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