Jason (jcreed) wrote,
Jason
jcreed

A strange question of just occurred to me, the answer may be obvious, I don't know:

Think about PA. Of course any consistent finite extension of PA is incomplete. Say a set S = {φ1 ... φn} of sentences is special if for any i, we have
¬(PA ∪ (S \ φi) |- φi)
and
¬(PA ∪ (S \ φi) |- ¬φi)
Zorn's lemma says there is a maximal special set Smax, and so all the 20 models of PA are given by all the different ways you can satisfy or not satisfy each of the sentences in Smax, because of the specialness condition (all the sentences are independent).

So think of the standard model N, and consider the bitstring we get out by reading off which elements of Smax it satisfies. Invert that whole bitstring. We have obtained a sort of "opposite model" of the standard, relative to PA, and relative to our choice of Smax.

Or, and this is the question I'm asking myself, is it really relative to the choice of Smax? Alternatively, is there a uniquely determined "opposite" model that disagrees with N on every sentence independent of PA?
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