[FOM] Reply to Neil Tennant
steve newberry
stevnewb at ix.netcom.com
Sat Feb 8 17:49:36 EST 2003
{Reply to Neil Tennant's crushing refutation of my inadequately formulated problem.]
Of course you are right. I WAS thinking in terms of General Models -FOL semantics, and should have made that explicit. Clearly, I **must** be more explicit, and henceforth shall do my best in that regard.
[read '@' as the epsilon of membership, ' /@ ' as its negation, and ' /= ' as "not equal".]
The logic is classical, simply-typed, of finite order, FOL-General Models syntax and semantics, and the ground domain is fixed, discrete and countable. It either IS omega, or its elements are indexed by omega. Under those preconditions, I ask you to consider the partition of all cwffs [of the above stipulated logic] into two negation-invariant blocks: U and NK, where a cwff B is said to be **absolute** iff it is either a contradiction or a tautology; and B is said to be **contingent** iff it is not absolute, hence NK is the contingent block. But NK splits very nicely in two, to give N + K, so we end up with three negation-invariant blocks: U + N + K, where B @ a block <=> ~B is also @ that block. Then
B @ U <=> B is absolute, i.e., ~B is a contradiction and has no models of any cardinality, and B is the negation of a contradiction, which, for lack of a better word, we'll call a tautology. B is "u-valid" [universally valid]
B @ N <=> ~B has an infinite, but no finite models, and B is valid on all and only finite domains; ~B is asymptotically close to being a contradiction, and B is asymptotically close to being a tautology. B is "n- valid" [valid on all and only domains of cardinality n, where n @ omega.] N is not r.e., and is the reason for the "undecidability" of classical logic. Tarski said it better: Classical logic is omega-incomplete. [N is a very interesting set of propositions.]
B @ K <=> both B, ~B have finite models. The models of cwffs in K are either finite and co-finite, or infinite and infinite, which I call "co-infinite".
I claim that this partition , negation invariant and defined by the cardinality of domains of realization, is complete, i.e., every cwff is accounted for, and falls into one and only one of the blocks.
Are we okay on that? GM-FOL semantics being now understood. If not, shoot me down. If so, then where do you stand on the paradox?
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I said: "A question which has nothing to do with predicativity:"
I was wrong about that. I am so locked into the predicative ontology that it is only by an effort of conscious will that I can do IMpredicative logic, and so the GM-FOL semantics is the way I think; whether I'm conscious of it at the time or not. This is by way of apology for wasting your time by not having specified the frame of reference beforehand.
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The "Paradox": [This is not a real paradox, but merely somewhat counter-intuitive.]
Suppose we have a cwff ~B which has an infinite, but no finite models. [It can be the conjunction of the axioms for any theory that has no finite realizations.]
Then B is n-valid. If B is **n--valid** then B **has models on all finite domains**. The union of those models would have to be an infinite model. [I'm not invoking compactness here, merely the existence of unions.]
NT Q: The union of those models would have to be an infinite model of *what*, exactly?
SN A: An infinite model of the n-valid cwff B of which I was speaking. I'm suggesting that an n-valid cwff is a cwff which **by merely existing**, in effect says, "the cwff of which I am the negation is satisfiable but not valid on omega, hence by the **existence** of the union of all my finite models, taken together with the fact that that union is necessarily infinite, I must have an infinite model, despite all my protestations to the contrary."
Respondez!
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