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Metaphysics
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 Equality In The Trinity So then, we actually have a third member of the trinity, the Holy Spirit. Whether you imagine the group C7 or the "universal being" that reasons so perfectly that it may override the virtues of God to the benefit of another, we should admit that the example of Jesus Christ towards us must indeed stay the same forever. That is; there will be no further requirement of commandment above the law together with the holding of L(G) intact. (ie a set of virtues must be continuous) That (G&p)~H in conjunction with H~(G&p) is greater than or equal to some G~H and H~G would show that the trinity in some form will attain some set of virtues, those being positive properties that do nothing to private God of any positive properties. We label both the complete set and elements of these virtues "q". Then it follows that as there is some maximal set of q, then they entail positive properties from those virtues. If these entailed positive properties are themselves virtues we are satisfied as we may see alternatively that if they were not virtues then the property which entailed them could not be a virtue. i.e. q1=>p, p=>¬q2 would entail both q1=>¬q2, thus q1 is not a virtue. However positive properties may therefore private other positive properties (possibly even a virtue) although a virtue may not private another positive property: i.e. P(p1)=>¬P(p2) follows from P(p2)=>¬P(p1) However if p1=>¬q we have a contradiction on modus tollens: but then p becomes not necessary. (and we assume N(q)) so that N(q) => N(p) So if the set of all such p are merely possible positive properties We would infer that as they are not virtues, i.e. N¬(q&¬p) then q=>p but modal status is always necessary so we may have N(q=>p) from which we gain N(q)=>N(p) if and only if ¬N(p)<=>N(¬p) but as p is merely possible, and is not necessarily false, we can not have q=>p. Thus the set of q is closed. Clearly q1=>¬q2 is not possible for a virtue (by definition). So the set of possible "p" may not logically follow from a virtue. However if it is true that p=>¬q for some virtue then q=>¬p or N¬(q & p) and such a positive property may not exist alongside that virtue "q". Rather than dismiss "p" as irrational, we willl do the alternative: that with the inclusion of p some virtue(s) q is excluded from the closed set of all such virtues.(**) Likewise positive properties may not be such as to agree with each other (although both are positive and so is their conjunction if there be one (if there is not, then only one is truly positive as they would be exclusionary)). We will state then that if two positive properties are in agreement with the same set of virtues, then they are in agreement with each other. Clearly if q as a set is closed then if p1 and p2 (not virtues) may not both hold then it is clear no virtue has entailed them both, but we will state that if they conflict with each other, then they do so in a manner that N(p1&¬p2) that is, they may not entail each other: One privates the other - they can not each be virtues. In fact we have P(p1) <=> ¬P(p2) Assuming they are both not virtues we would then be able to state that since they exclude each other and are not entailed by any virtue, then if N¬(q1&p1) say, for some virtue q1 as per the above (**). Now if from the definition of a virtue the set of all q can not private either p1 or p2 then neither will a subset of those in q. Since p2 would also therefore private some virtue q2 we have that the set of all virtues without {q1,q2} neither privates either of those properties p and p2. However p1 still seems to private p2. However , as pos(p1) and pos(p2) it can not be true that they are opposites: rather the set of virtues q has conflicted with these sets of positive properties in p. Then {p1, p2, {q}{q1,q2}}, the set of remaining virtues now may be said to include p1 and p2 if q1 and q2 are excluded. How do we show this? We have ensured that P(p1)=>N(p1&q) holds with P(p2)=>N(p2&q), so if P(p1&p2) then N(p1&p2&q) Assuming ¬P(p2&q) and P(p1) then N¬(p1&p2&q) so that p1&p2 => ¬q (a contradiction, we excluded all such q before, we are already "in lowest terms" and q is assumed to not be empty - if it was then P(p1&p2) immediately is possible and results in N(p1&p2).) Therefore; p1&q and p2&q both hold. Which if P(p1&p2) then {q, p1, p2} is now closed if p1 and p2 include all those terms which are entailed from them. (for if p1 entails say, p[n] then (p1 & p2) => (p[n] & p2) etc. And no such p[n] are privated by any q or therefore anything following from p2 if P(p1&p2). Therefore our reduced set of virtues is now reduced and p1 and p2 grafted on. We have a new set of virtues! Therefore assuming some set of virtues q0, we may shift to a different set q1, q2 etc and our trinity is "dynamic". We ensure the exclusion of 'previous' virtues leaves a set "q" intact, we will like this to the strictest container, Jesus, and the set of every possible positive property of all possible (not just compatible) p and q to be the Father. Then, as we can see from our model using multiplication on K4 subgroups of the octal, we liken "q" to the intersction of two (or three is the maximum) K4 subgroups (there is always an intersection of pairs of K4 groups in the octal) and q with {q1,q2,..} or the new elements {p1,p2,...} as in the remaining slots in our K4 group. The K4 groups of the octal cycle with multiplication, We with our new set of virtues have say, {p1,p2,...} as alike to 0 = the octal, (everything) = (0,a,b,c,d,e,f,g) Where we liken 'a' as unity to [a,b,c] so b=[a,d,e] and c = [a,f,g] If this last paragraph looks strange to you, you will need to become familiar with the content of the metamath and revelation sections. Then under frobenius in the octal inducing multiplication on the K4 ultrafilter above, and then in any place in the Father by multiplication, we may see the structure is actually immersive. All we are required to show is that the sets [a,d,e] = q,p1,p2,... and [a,f,g] = q,q1,q2,... obey some logic. We now know under a form of addition in the octal; two K4 groups my be added so A+B = (AvB)c where the complement is taken in the octal. ('v' is the symmetric difference). So we actually have in the sets above, {q,q1,q2,..} + {q,p1,p2,..} = (everything) - {q1,q2,...} - {p1,p2,...} = q as desired. Of course, we are using very simple set notation here, but for all intents and purposes we are actually using operations on an infinite and non-principal ultrafilter. The set q, and its supersets are plausibly infinite. (it makes sense then to use "everything" in the operation!.) However if q is infinite, then it may be also co-finite we may also have a finite set in the q[i] and p[i] also! So, we retain some sense of a finite set of requirements which are pleasing to God that may also be required of us. Those that are always present perhaps. If the intersection of every possible 'q' corresponding to unity is in some way spanned by the law for example: We would be very well blessed indeed! We know that the set q belongs in the ultrafilter (by construction) so therefore do the supersets in q[i] and p[i]. In effect, we have our extension in completeness, yet there is a little more to follow! Then if we act with multiplication with the valid seven cycle (a,d,f,e,b,c,g) on the K4 ultrafilter above, then; 0 = the octal, (everything) = (0,a,b,c,d,e,f,g) becomes And we see that there is some validity in linking the previous state to the next under multiplication: For the group [a,d,e] in the former is associated with the case after a transformation, whereas after the multiplication by the seven cycle it becomes the next case under transformation. (i.e. the p's become the q's for the next time.) No matter the subgroup and the choice of unity, the structure of the octal will ensure there is always such a case. |
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