Given All The Facts
In every subgroup containing virtue it is simple enough to state that the conjunctions of r and s etc. are empty and not in the indexing set. But with the four subgroups that do not so contain virtue; does the indiscernibility of positive properties from a virtue suffice to close the sets? I argue that yes, it most certainly does. Each of the six properties not virtues are constructed in the maximal sense as in predicates of sets; being all "a" such that a=>r or all "b" such that r=>b etc.
In these properties it is easy enough to argue that these equivalents are "indiscernible enough" so that the statement that for instance, given v, I have the K4 triple r + u&v = v.
I must argue that r&u&v is not in the indexing set.
Now in the book it is argued that r and u&v are disjoint sets. Then the intersection is empty. That is not to state that r&u&v is not a positive property, is it?
If Pos(r&u&v) then N¬(Pos(r&u&v)^{-1}) <=> N¬(r^{-1}&¬(u&v)) which is also to state that; r^{-1}=>u&v or (u&v)^{-1}=>r.
Then the positivity of the excluded middle implies that the disjunction r v (u&v) holds, but then there is no excluded middle. So the inconsistency of Pos(r&u&v) is shown. it may not be a positive property: it privates itself. (There is also no equivalence to virtue, one property must be axiomatically "positive". The statement should be properly formed: I.e. r v (u&r^{-1}=> u&v). Then N¬(u&r^{-1}&¬v) or N(r&v&¬u) and Pos(r&u&v) is "impossible".)
Likewise given Pos(x&y) I have "x v y" in every sense it may be applied: The facts are such that it simply holds that each triple from the octal which together form subgroups are also closed.
Then what may I state of Pos(x&y)^{-1}? only that Pos(x^{-1}&y^{-1}) holds and so one at most of x and y holds as positive. Yet once again this follows from the disjunction on both sides. Then because x&y is not a positive property, (x&y)^{-1} is not exemplifiable in a positive sense as given the same disjunction, (x&y)^{-1} is or must be "out of scope". I should limit the indexing set again.
In fact given p, the 'negation' of (r&s)^{-1} is not r&s but is actually u&v. If I am limited to the indexing set {p, r, s} and r&s is certainly not in the filter then it does not follow its complement must be! This complement would be in I \ W for some W in U, so must be in the indexing set: but I modified the indexing set to exclude r&s and its inverse! There is no quandary!
Then x&y is not positive, and may never be so, then what of its negation (x&y)^{-1}? In fact, all I may state is that given free exercise between x and y, x&y is negative. (may never be positive.) There is one fallacy of equivocation to watch for and that is for (x&y)^{-1} to be mistaken for a virtue. (But it is entailed from both sides of r and s).
So then, every triple forming a K4 subgroup of the octal has a different indexing set: one that does not contain whatever analogue of x&y and (x&y)^{-1} is not "in scope" (in the indexing set) and also the union of these indexing sets is the whole set of positive properties - the ultrafilter of the octal group itself. It would almost seem that a lack of free exercise between Pos((x&y)^{-1}) and Pos(x&y) is enough to exclude (x&y)^{-1} from the indexing set.
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