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Polyps:
Strongs:  Metaphysics Infinity And Incompleteness Given I had split the sets of A and B (A = {r : q=>r} and B = {s : p&q-1=>s}) into subsets where A is split into R={r : q=>r} and U={u : q=>u-1} and B is split into S={s : p&q-1=>s} and V={v : p&q-1=>v-1} I have somewhat related these sets by a virtue of "liberty". Now, "Liberty" as a virtue accounts for those sets in u and v which may become consequent of filtering from the set of A and B those predicates free to be non-exemplified in any such freely decidable disjunction. Now, if I maximise all R to A and S to B, then U and V are "empty". This has the problem that, as in chapter two of the book there is yet the quite separate and existing positive property of (r&s)-1. This is separate in the set of all positive properties from both A and B. What happens to the rest of the octal when R is maximised to A and S maximised to B? The disjunction (r&s)-1 v p&r&s => u&v&e=>e appears to only imply "e" (as u&v is "empty") Then if u, v (and therefore u&v) are all empty, I may rearrange to (u&v) v p&(u&v)-1 => (r&s)-1. Then as liberty acts on one side only I necessarily have the right hand side, i.e. then (r&s)-1 is necessarily positive. Then the octal is incomplete; and unless "s" is empty (i.e. R is maximised to the set of ALL positive properties i.e. (r&s)=R\e), there is immediately found an inconsistency, the octal whole may not be formed on the basis that some positive properties, though they be in appearance required to be necessarily rested upon, may then not become properties exemplified, also by a consequence of liberty in virtue (a "u-1" becomes an "r"). So, if I find liberty restricted; in not being able to rest on every r in R (so U is maximised to A), or as before unable to maximise R to A, Then I find the sets of the octal must be necessarily non-empty. There appear to be some "indispensable" properties in each set. Now, the octal as a result was based on the fact that q and p as positive predicates both entail closed subsets: This implies that there are other positive properties in the set of all positive properties in the consequent middle r&s as well as in u, v and u&v. If there are always such sets, then I discern that either God is truly infinite in positive properties or somehow the equivalence of N¬(r1&r2&¬r1) to N¬(r1&r2&¬(r1&r2-1)) ensures the sets u and v may ALWAYS without fail become (as by choice) non-empty as I will always have a partition of A into some non-empty R and U, i.e.... there always exists such a partition (unless A only has the one possible member). However, I am interested in God as infinite, so I will state that every set of the octal necessarily has one member as it is "God" and equally so because "God" is necessarily existent. I immediately infer that (r&s)-1 is necessarily positive, closed and without equivocation, effectively as "zero", as the other sets of the octal u, v and u&v are "empty". I have by equivocating (r&s)-1 with "e" inferred some "morphism" or injection into {e, p, r, s}. I have a sense where the octal in some way self-references the only K4 group remaining in that set that should be the octal. Now if each of the seven sets of the octal are always non-empty: Not only MUST I have some predicate in both u and v, I may equally find there is of necessity an octal group which is an arbitrary re-arangement of A=R, B=S into R and U and S and V. I have this from the equivalence of N¬(r1&r2&¬r1) to N¬(r1&r2&¬(r1&r2-1)). However I required that R contains some r1 and also U then also contains some r1-1. Are these sets legally "disjoint"? Will they also form an octal? r&u-1is just r1, is this a sufficiently positive statement to satisfy and give a non-empty octal, with no empty "sets"? Clearly I only need to examine s&v-1 which is just equal to s1 similarly: (u&v) v p&(u&v)-1 => (r&s)-1 is basically (r1&s1)-1 v p&r1&s1=>(r1&s1)-1 and the statement is necessarily decidable as before, even the disjunctions not in virtue suffice to show that there is agreement; for instance: Given v in the place of virtue, v&r v (u&v) which is in this case: (s1-1&r1) v (s1&r1-1)=>(r1&s1)-1. This also makes sense, as do the other rows of the filter table of chapter four by symmetry). Now, that is truly a "fudge", but there is a simple rule that the local closure of K4 group is never equal to the full closure of the octal. Then, God is properly found to be "infinite". For then, no matter the disjunction, the result is always a "closed set" and forms a locally closed K4 group as in virtue. Then as by assumption there is always an octal and by that assumption every set in it is necessarily non-empty, it is impossible for the local closure to be equal to the closure of the octal, even though they may appear to self-reference each other. Why? well, more faith than anything else, yet I still find that by extending the octal over the local closure, every possible disjunction that forms a K4 group will in the greater and containing octal (a degree of extension over it), will always imply some non-empty U and V from A and B, despite R and S being locally maximised to A or B. A and B may be "maximised" but only locally, and the octal is always complete over it: and I assume then that there is always a greater set of U and V to be found and so by some self-reference I simply state that there is always such a U and V in the greater octal present. So, no matter the choice of q, there is always a mapping table - and also a filter table, and therefore always an octal. Every choice of q is to generate a closed set in the octal, so then no matter what, the other sets in the octal exist and are non-empty provided the octal of all positive properties is properly chosen in "completeness". Continue To Next Page Return To Section Start '