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thanks

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- Thread starter wisvuze
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thanks

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AKG

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In other words, the question is whether the Henkin structure associated to [itex]\mathrm{Th}(\mathbb{N})[/itex] can be (isomorphic to) [itex]\mathbb{N}[/itex]. The answer is Yes. In the Henkin construction, you add [itex]\omega[/itex] constants to your language and [itex]\omega[/itex] axioms to your theory. You then repeat this [itex]\omega[/itex] times. You then extend to a complete theory. And then you construct the model as the set of variable free terms in your new language, modulo being provably equivalent by your new theory. What you need to do is make sure that when extending to a complete theory, for every Henkin constant c there is some "SS...S0" such that "c = SS...S0" is added to your theory.

This is easy: Look a the first set of Henkin axioms you added, they're of the form [itex]\exists x \phi (x) \rightarrow \phi (c)[/itex]. If [itex]\exists x\phi (x)[/itex] is true in [itex]\mathbb{N}[/itex], say [itex]n[/itex] is the minimal witness, then add [itex]c = \bar{n}[/itex] to your theory in the final extend-to-a-complete-theory stage. Here [itex]\bar{n}[/itex] is [itex]n[/itex] S's, followed by a 0 symbol. If [itex]\exists x\phi(x)[/itex] doesn't hold, add [itex]c = 0[/itex]. Now deal with the Henkin axioms added in the second iteration in the same way (by looking at whether the existential sentence is true in [itex]\mathbb{N}[/itex]), interpreting any occurrence of the first set of Henkin constants according to the interpretation we just fixed above. Do this for all the Henkin constants/axioms, and then complete the theory as usual.

It's not hard to see that this will be isomorphic to [itex]\mathbb{N}[/itex].

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Thanks for the reply, I appreciate it

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