L(G) Has Correctness

An ultrafilter is "true almost everywhere" in the model it is used in. We saw how the set HG(G) \ L(G) did not imply that a believer could hold this set of beliefs and then commute a transformation in the virtues of God under his own liberty L(x). (That is unless he is the individual G). Thus the model is indeed true almost everywhere (but correct). The result is the statement, "there is none good, not one - all have sinned and fallen short of the glory of God" is truth enough that there are none in the set {x} that would lay claim to HG(x) = HG(G) \ L(G) without L(G=x) (Only Christ may do this).

However, we should give further thought to the possibility that there may be two sets of virtues applicable to us rather than just the one.

Clearly L(G) would require 'x' to submit to the whole set of commandments. Is it possible for two sets of comandments to be valid, both with L(G) intact? Immediately, considering the maximal sets HG(G) \ L(G) and JG(G) \ L(G) as these two sets of positive properties we may state from the ultrafilter that in the non-principal ultrafilter of perfections {HG(G) ∩ JG(G) } \ L(G) must also be in the ultrafilter.

In assuming that these two sets are not equal, we must have that their intersection is strictly not empty: unless it be the case that the structure is actually positive enough for the two sets to be consistently disjoint: and then we may infer some form of morphism or bijection between the two classes. That would give some "well-definedness" to the model.

If the sets are equal we are done, however let us assume that the corresponding sets q1 and q2 of virtues are not equal. Then their intersection q should not private any property in the union: So we must have in some minimal case that virtues p1={q1\q} and p2={q2\q} private each other over q. (As they in q are all "core" virtues. However assume that q1 and q2 are not virtues, but are merely positive.) We will assume that since there has been a transformation, shifting a set of virtues has left us with two apparent sets of virtues, whereas q1 and q2 are "not virtues" as q alone was previously.)

Therefore {p1,p2} is plausibly positive over some reduced set q0 a proper subset of q (the intersection). In both cases we have L(G) - that it does not private any value in the set {p1,p2,q0} This set is closed: however the intersection q was such as to allow p1&p2 to private each other. So therefore if there is a commandment to preserve the pairing of these two properties p1&p2 there must technically be some fault in a virtue in q\q0 so we just state for now that p1&p2 => ¬P(q\q0)

Now given q, P(p1)<=>¬P(p2)

Now (q\q0) are all virtues and can not private p1 , p2 or p1&p2

But p1&p2 can not be a virtue since it privates those virtues in q\q0

and we also have N¬(P(q\q0) &p1&p2)

But if there is some fault in a virtue in q\q0 it is not intrinsic: q\q0 would indeed be a coherent subset of positive properties not privating any other. (it being formed of "core" virtues and not added to or taken out from the transformation of the trinity.)

So we assume the converse; that there is a commandment to preserve q\q0 with a fault in p1&p2 Then one or both of p1, p2 together are not "positive": one at least will break the commandment preserving q\q0, and will therefore break a commandment to preserve the wider set of virtues q.

So N(q\q0) would entail that ¬P(p1&p2) given q.

So, every set of fixed virtues say, q1 and q2 is either totally distinct (q=q0) or there is some fault in the law rooted from ¬P(p1&p2)

The former case q=q0 results in p1&p2&q being coherent and we would infer that p1 and p2 simply do not private each other, but then the union of q1 and q2 is equal to the union of p1 and p2with q, so the sets q1, q2 are equal. (They are completely compatible as virtues together. Therefore every set of virtues must be totally disjoint) Or, else there is simply a commandment against holding p1&p2together, and therefore there is a commandment for not holding q1&q2. Thus the virtues are incompatible. (Thus any two maximal sets of virtues are incompatible.)

Now, that there could be two separate laws and that no two laws may be held together are not incompatible concepts, for if there is merely a commandment against mixing laws, then we have a whole vista of commandments in our possible qi

So, to wrap up,

HG(G) \ L(G) and JG(G) \ L(G) are therefore equal both to their union as one set of laws: or they are restricted to disjoint sets or "other laws" that are incompatible and separated by commandments. However then, as virtues there would be a set of commandments that are true everywhere such as L(G). That is, unless L(G) is a principal element. Alternatively L(G) could be an "essence" of this ultrafilter.

Of course N(q\q0) => ¬P(p1&p2) which is otherwise p1 => ¬p2 ("core" virtues are maintained and p1, p2 are "right" and "wrong" respectively.)

So vacuously the statement q = q0 => P(p1&p2&q) a contradiction: (Laws are identical)

But if the laws are completely distinct and disjoint, then we have a contradiction, since both must at least contain L(G). Hence there must be only a single law.

Therefore there is a well founded principle for forming laws upon virtues that do nothing to restrict "core" virtues, but analogously the laws must prevent the shift of "core" virtues to another set. One core set of "pleasing" virtues is assumed, and there are commandments in order to block the methods by which those sets may be altered. We could extend with many positive p1's, but with commandments against many p2's whether they merely conflict or are truly negative Thus, we have "right and wrong" upon a set of virtues.

Of course, as our creator God would not issue laws that we were not able to break ourselves, even if we were to try! He would not "command us not to fly when he had not given us wings" etc;

Therefore the virtues chosen in L(G) have some framework of correctness! (I.e. One law only, with consistency and agreement in the set of further positive properties upon a core set of virtues.)

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