I need to prove this,

Let $T$ be a theory in language $L$, let $T'$ be a definitional extension of $T$ to language $L\subseteq L'$.

  1. If $\mathcal {M} \models T'$, then $\mathcal M$ is a definitional expansion of $M\upharpoonright L$.

  2. If $T$ is complete, then $T'$ is complete.

  3. If some model of $T$ has a definitional expansion to $T'$, then $T'$ is a definitional extension of $T$.

My problem is that it feels like all that is true merely by the definitions, so I guess I'm missing something, and would like your input about it. The definitions I'm working with are:

An L'-theory $T'$ is a definitional extension of an L-theory $T$ if $T \subseteq T'$ and $T'$ is axiomatized by $T \cup \Delta$ where $\Delta = \{\delta_s | s \in L' - L\}$ and each $\delta_s$ is a definition in $T$ of the symbol $s$.

A definitional expansion of $M$ is an expansion $M'$ of $M$ to some language $L \subseteq L'$ such that for each symbol $s \in L'- L$, $s^{M'}$ is definable in $M$.

  1. Let $\mathcal {M} \models T'$, and $T' = T \cup \Delta$ where $\Delta = \{\delta_s | s \in L' - L\}$ for each $\delta_s$ of each symbol $s$, so it looks pretty clear that that is exactly the definition of definitional expansion. Any ideas about how to argue it better?

  2. The idea is, $T'$ and $T$ prove the same things, the extensions are conservative, clearly if $T$ is complete so will be $T'$, but I have trouble saying it formally enough, or did I miss something?

  3. Same issue it feels like just 2 different shades of the same definition, and I don't know how to connect them, besides noting that we are talking about pretty much the same thing.

I would very much appreciate your advice, I feel slightly baffled.

  • 1
    $\begingroup$ I think "let $T'$ be a definitional extension of $T$" only applies to statements 1 and 2. The statements are true not "merely by the definitions", you need to use the soundness and completeness of first-order logic, but given that the proofs are quite simple. $\endgroup$ – Rob Arthan Apr 19 '15 at 9:34
  • $\begingroup$ Actually, for 3 $T$ is a complete Theory and $T'$ a complete extension, you are right, I missed that. I'm a bit confused at how to use soundness and completeness here, could you please help give me some advice about it? I do understand that what I'm trying to prove is the link between semantics and syntax and that is exactly what completeness do, but I don't know how to start the proof. Thank you for your help. $\endgroup$ – Sara Apr 19 '15 at 9:48

As to 1., indeed, it isn't too difficult. Just choose $\delta_s$ to make $s$ definable in $M$.

As to 2., it's just: $$T' \models \phi \iff T' \models \phi^L \iff T \models \phi^L$$ These equivalences deserve some extra attention. The first is just a substitution theorem, the second requires you to prove that every $T$-model can be extended to a $T'$-model. Then we can apply completeness of $T$ to deduce completeness of $T'$.

Finally, for 3., denote the model with $M$, the expansion with $M'$. Since the new symbols are definable in $M'$, we can use the defining formulas as a hypothetical $\Delta$ (this works because $T$ is complete). To conclude that $T'$ is actually axiomatised as $T \cup \Delta$, we will need to use the other completeness hypothesis.

As requested, some more detail on the second part. We need to prove that:

$$T'\not\models \phi \implies T'\models \neg\phi$$

Suppose that $M \models T'\cup \neg\phi$. Then $M \restriction L \models T$. Then by completeness of $T$, $T \models (\neg \phi)^L$. Consequently, $T'\models (\neg\phi)^L$.

Note that we only required the "easy" direction:

$$T\models (\neg\phi)^L \implies T' \models (\neg\phi)^L$$

which amounts to the fact that every $T'$-model carries a natural $T$-model.

I guess that would've sufficed.

However, there were other things at play here which came along quite naturally -- at least to me.

Namely, the converse is of considerable conceptual importance. Suppose we forgot about $T$ being complete. Here it is once more:

$$T'\models (\neg\phi)^L \implies T \models (\neg\phi)^L$$

What does extending every $T$-model to a $T'$-model mean? It's not hard, really (define each term $s$ by means of its $\delta_s$, prove by structural induction that the result models $T'$). But why bother?

If we pass to the proof-theoretic side of things, we obtain:

$$T' \vdash \phi \iff T \vdash \phi$$

That is to say, all these fancy new symbols, relations and functions are:

  • Not necessary: we can do everything in $T$ if we insist;
  • Permissible: nothing goes wrong by using the new symbols of $T'$!

Thus, if we choose to work in our favourite theory $\sf ZFC$, we can make our lives easier and stick to the mathematical practice of introducing new notation, using convenient shorthands like:

  • $x = \varnothing$ instead of $\forall y: y \notin x$;
  • $x = 2^y$ instead of $\forall v: v \in x \iff (\forall w: w \in v \implies w \in y)$.
  • $\begingroup$ Thank you for your input, I will work on this and come back in case I hit some other wall, but I definitely see things clearer now. $\endgroup$ – Sara Apr 19 '15 at 10:32
  • $\begingroup$ Hi, I'm a bit stuck in the second one, I'm having issues with the part about proving that every T model can be extended to a T' model. I get the idea in general, it is: T is complete, every T model can extended into a T' model, so I extend it, once its extended I substitute the things on the definitions, so it's still complete. But I don't really know how to go about it, or to say it in a clearer way, I would really appreciate your advice. $\endgroup$ – Sara Apr 19 '15 at 17:41
  • $\begingroup$ Thank you very much for your help, the formal aspect of this is way more clear to me now, not only the intuitive part. This has been really helpful! $\endgroup$ – Sara Apr 20 '15 at 10:53
  • $\begingroup$ @Sara You're welcome. It was also helpful to me to go through these things once more. Cheers :). $\endgroup$ – Lord_Farin Apr 20 '15 at 12:40

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