Power To Take It Up Again
Now given I may make the substritutions r = r1, u = r1^{-1}, s = s1, v=s1^{-1}, This immediately results in the disjunction (r&s)^{-1} v u&v being equal on both sides.
But, on maximising that set of "r" to all Ω and limiting the set in "s" to "e" alone, I have by rearranging Ω v e gained <e> v ∅. This is actually as true as the same disjunction in (r&s)^{-1} as above. (It is also decidable.)
Now, given our predicates were of "x", then stated "God exemplifies property 'x'", How may God freely decide to exemplify nothing?
Clearly His existence in <e> is necessary, i.e. "provable" if existent at all, yet in that same disjunction, the octal has collapsed to just one K4 group with that disjunction in (r&s)^{-1} being found as if "complete rest" on all Ω.
So, (r&s)^{-1} is effectively empty or as ∅=<e>^{-1}. This results in (r&s)^{-1} v u&v being null both sides. (Both sides are inconsistent!)
Then it is a simple observation to note that N¬(p&r^{-1}&¬s) still holds : and N¬(p&¬(r&s)) is simply a nonsense, virtues are closed and (r&s) is inconsistent given p and r or s.
So, as (r&s)^{-1} v u&v must be null both sides, There is no "positive property" outside of the K4 group formed in "p", "r" and "s".
Equally, I may then drop u&v and simply treat u&v as empty instead.
What does this mean for the octal?
In any such disjunction there is of necessity a middle of excluded positive properties, whether r&s or (u&v)^{-1} etc. There are always positive properties outside of the local closure: admitting the existence of (r&s)^{-1} permits the disjunctions of the filter table of chapter 4. I expect to find in r v (r&s)^{-1}&r^{-1}=>u^{-1} and s v (r&s)^{-1}&s^{-1}=>v^{-1} a solution for u and v from the closure of positive properties in the octal, rather than local to the K4 form.
This I find by axiom of virtue, and by the action of multiplication of C7 over the octal. "r" and "s" in these two disjunctions are found by "multiplication" of the previous disjunctions in "p" to an isomorphic octal where "r" and "s" are "become as virtue". This allows the axiom of virtue to find some r and s suitable for generating u and v. (I simply need by axiom that there is some set of u and v not in (r&s)^{-1} or, of course, r and s (or p).
May I, as if by axiom show there always to be such a "u" or "v"?
Clearly, the notion of infinity is apparent: For there to always exist such properties in a closure over the "octal" (then made local to <e> in the K4 group) one should extend from one infinite cardinal to another: There is no "most infinite" infinity. To assume that I may simply "choose" what is then always a generator of just one subset of seven such disjoint sets of positive properties to instead generate the complete closure of "everything" from it is not "definite", there are also the possibilities of infinity, and a solution for "God" is also such a solution for a "set of everything" - at least of positive properties.
So, what of God freely deciding to exemplify nothing?
We arrived at a possible contradiction that (r&s)^{-1} v u&v is empty both sides, that the octal may be inconsistent rather than necessary. Our contradiction was that the octal may be limited to r = r1, u = r1^{-1}, s = s1, v=s1^{-1} (as top). Then our initial assumption, that there are no positive properties in u or v outside of p, r and s (and r&s)) is at least, suspect. Of course, we also assumed the existence of an octal too,... there is possibly no octal to be found over the K4 form when maximised, but that r&s exists is completely certain unless r = <e> and s=∅. For then, God is free to limit Himself to any Ω or r&s is simply become a rearrangement of r v s.
So, God is free to rest, and the middle <e> = r&s becomes logically necessary: and God in that middle is "perfection existent". That God Himself is free to rest on solely His existence; as s = e entails that <e>^{-1}= (r&s)^{-1} is as non-exemplifiable as <e> or r&s in that disjunction
So, In freely exercising e v Ω I find that the middle is necessarily empty: yet the middle is "of everything" for three K4 groups out of seven.
Consider the verses:
Joh 10:17 Therefore doth my Father love me, because I lay down my life, that I might take it again.
Joh 10:18 No man taketh it from me, but I lay it down of myself. I have power to lay it down, and I have power to take it again. This commandment have I received of my Father.
(KJV)
Now, in this octal (with u and v empty), three sets of seven (three days in the grave) are null or void. In Christ laying down His life by exercising Ω v e There is the freedom to rearrange the disjunction to <e> v ∅ despite <e> being part of the impossible (non-exemplifiable) middle. Christ has the power to take His life up again. I also find the verse:
Mat 27:46 And about the ninth hour Jesus cried with a loud voice, saying, Eli, Eli, lama sabachthani? that is to say, My God, my God, why hast thou forsaken me?
(KJV)
...And that there should always be found an octal over Him: (c.f. John 10:17 above.) Now, given (r&s)^{-1} and u&v equal and so inconsistent, the octal has "departed" from the K4 form, u, v and u&v become empty, and afterward Christ exemplifies "nothing" for those three "days". Finally rearranging to <e>, God is become (once more) perfection existent with the full octal regenerated in its complete closure with the seven cycle in place as acting upon <e>.
There can be no solution for God in triune form without both the octal in place and then always outpacing it's own closure by some form of self-reference. I don't have this answer: merely a best guess.
There is another trick to be found of rearrangement: r v (r&s)^{-1}&r^{-1}=>u^{-1} and s v (r&s)^{-1}&s^{-1}=>v^{-1} in the octal may become sound by putting u=s and v=r. Then s=r1^{-1} and r=s1^{-1}. This is also the rearrangement of (r&s)^{-1} v p&r&s=>u&v to simply (r&s)^{-1} v r&s.
This extra substitution has no "new" property other than when maximised, results in either <e> v ∅ or e v Ω. Liberty also, is present but found only in entailing "rest" by axiom of virtue in this case.
The K4 form then, when maximised may simply regenerate in a disjunction r v s that disjunction of e v Ω, similarly u^{-1} v v^{-1} is also the same disjunction e v Ω. However, the third group (r&s)^{-1} v r&s is actually <e> v ∅. So, on the "third day" Christ is by necessity "risen"; when the former two from the octal are "added" by using the complement of the symmetric difference.
I once more must assume an octal in which to make the addition.
Jesus was perhaps talking metaphysically when He said "on the third day I will raise it up again" as opposed to being "in the grave" for three days. (As different durations.)
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