Metaphysics

Local Closure In The Octal
So, given that every subgroup of the octal is now closed, It is a simple observation that the group C7 also indexes the singleton elements from the predicates of sets in the octal whole. Mapping singleton to singleton is preserved as well as mapping subgroup to subgroup and also the whole set of positive properties that may be indexed by the octal (and the ultrafilter) itself.
If you wonder why the classically positive property r&s is not "a positive property" in the octal  not even being in the indexing set then simply repeat the whole application of the octal with q=r&s. There is no fault in the octal; merely that it is not logically possible to exemplify a logically impossible action if virtue is constrained, or if God is not free or able to act in a manner that is not possible without rewriting the middle of every disjunction.
Now, given that fact: the same octal structure holds whenever q is the conjunction of a set of positive properties  which is also a set in the power set of the set of ALL positive properties. There is then a clear pattern emerging: God, may rest and also be dynamic  God needs not change but may act perfectly on a small set in q as well as all q: and has no question to resolve concerning His actions or lack of them. He may even do nothing  but rest perfectly with the action of virtue and still be God.
I argue that the octal despite the "illegalities" of properties such as r&s is unconstrained. r&s are not so much identifiers as labels in place; and in one case an r may be a q or a v etc. In truth the union of the indexing sets is as positive as is required. Two indexing sets (K4 groups in the octal) always share such a label: whether a "p" or an "u" etc. there is always an intersection and the octal group ensures that their union is the set of all positive properties. Whilst r&s may also be treated as a "q" in another disjunction, it is the case in the schema that (r&s)^{1} is yet positive despite not being present in the indexing set of [0, p, r, s] (Which is closed.)
In fact (r&s)^{1} appears comfortably in the group [0, p, (r&s)^{1}, (u&v)] and r and s are not referenced in that filter's indexing set either!
Then it is a simple matter to extend the operation to subgroups. It should now make sense or be trivial that [0,p,r,s] + [0,p,u,v] is meaningful and is equal to [0,p,(r&s)^{1},(u&v)]. I take the operation (A v B)^{c} on their indexing sets and the result is immediate. (The correct labels are already in place with the same operation on the octal group's subgroups themselves.)
One problem remains, the identity element 0 is the set of all positive properties. Clearly using the operation (A v B)^{c } it suffices to show that whether on a singleton representation of a set; 0 + p = p as well as (p, r, s, u, v, (r&s)^{1}, (u&v)) + (p, r, s) = (p, r, s) etc. I simply need the indexing set of each triple as zero to meet the local closure of each subgroup in the octal. {p} would contain all virtues only; and so forth. The indexing set {p, r, s} of [0, p, r, s] would contain only virtues along with the positive properties that are possibly exemplified from either side of the disjunction, excluding those properties that are entailed from the middle, taken both sides of the same disjunction q v p&q^{1}.
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