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We say an $\mathcal{L}$-theory $T$ has built in Skolem functions if for all $\mathcal{L}$-formula $\phi(x,\bar{y})$ there is a function symbol $f$ such that $T\models \forall\bar{y}(\exists x\phi(x,\bar{y} )\rightarrow \phi(f(\bar{y}),\bar{y}))$.

$ \textbf{Question}.$ Let a theory $T$ have built in Skolem functions. How can we prove that $T$ has $\forall$-axiomatization?

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closed as off-topic by Leucippus, user91500, E. Joseph, Namaste, Daniel W. Farlow Nov 4 '16 at 16:14

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  • $\begingroup$ I think you are referring to the idea of "Skolemization" to rid the axioms of existential quantifiers. See for example Skolem normal form. The details of your "built in Skolem functions" are needed if you intend to "prove" $T$ has universally quantified axioms only, but the idea presented in broad terms is clear enough for most purposes. $\endgroup$ – hardmath Nov 4 '16 at 7:36
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    $\begingroup$ @hardmath The OP is referring to the model theoretic notion of definable Skolem functions, see here: modeltheory.wikia.com/wiki/Skolem_functions $\endgroup$ – Alex Kruckman Nov 4 '16 at 14:50
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You just need to combine two facts to prove this:

  1. If a theory $T$ has built-in (also called definable) Skolem functions, then every substructure of a model of $T$ is an elementary substructure.

  2. A theory $T$ has a universal axiomatization if and only if it's class of models is closed under substructure.

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