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 those conjunctions, as positive properties compared to 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-1.

I must argue that r&v-1 and sets like it are not in the indexing set.

Now, how do I exclude sets as (r&s)-1 in the disjunction of r∨p&r-1=>s in the middle using the operation of (a∨b)c? The answer is simple; I don't, the creator already has excluded those sets with the world (creation).

It is then the case that the middle in the disjunction will exclude the sets of r&s and u-1&v-1, and the disjunction of r∨v-1 and s∨u-1 are likewise affected by the middle but in a far simpler fashion (as to derive). The virtue is actually as u&v and not some axiomatic virtue in the set of "p".

For instance, beginning with v-1∨v I can expound upon the disjunction with the implicit binding of virtue in p, fixing the disjunction in virtue, without explicitly declaring it in the disjunction itself.

v-1∨v&(u&v)=>r, as desired.

That those works in u&v (which are already instantiated in a work going before) may be freely rested upon is a constraint of nature on the disjunction itself (inherited, as it were from the state of creation). They are enough to constrain the middle in the disjunction. The set (r&s)-1, however is not yet instantiated and would decide the disjunctions of r∨u-1 and s∨v-1 which have no excluded middle in the disjunction as yet - nothing uninstanced may constrain a middle, can it?

Then writing s∨s-1&(r&s)-1=>v-1 is a remainder, and I also have r-1=>s&v-1, so (r&s)-1=>v-1, also. The disjunction follows as required. (Note that only u&v can separate r∨v-1 as there is always an excluded middle!)

So, the sets (r&s) and u&v certainly act as virtue; as the two "keys" to open the octal schema from a disjunction with the binding of virtue, as derived from the "mapping" between the two sides formed only on the disjunctions formed on the virtue(s) in "p". They are determined uniquely and form a most "compact" solution, with no other sets in the disjunction apparent, and none needed. Then there are a minimum of seven sets in the schema, with a mystery of a zero to include also.

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