Skolem Theorem private case, preserving extension Question:
Let T be a theory of statements over a signature $\sigma$ and $\phi$ is a formula over $\sigma$ s.t. $x,y$ are the only free variables. We define a signature $\sigma'$ by adding a new binary function sign $f$ and a new constant sign $c$ to $\sigma$ .
Denote $T'=T\cup \lbrace \forall x \forall y (f(x,c)=y \rightarrow \phi)\rbrace$. T' will be called a preserving extension of T if for every formula $\beta: T\vDash \beta \iff T'\vDash \beta$.
Prove or disprove:
if $T \vDash \forall x\exists y \phi (x,y)$ then T' is a preserving extension of T.
My proof
One direction is trivial so I will focus on the more difficult side for my question: 


*

*Assume that $T'\vDash\beta$.

*Let M,v be a structure  s.t $M\vDash T $ so $M,v\vDash \forall x \exists y \phi (x,y) \iff \forall a\in D^M \exists b_a\in D^M : M,v[a/x,b_a/y]\vDash \phi(x,y)$

*(We now extend M to M' over \sigma' by defining $f^{M'}(a,c) =b_a$ and defining c as some constant in $D^M$.)

*$\iff \forall a\in D^M : M',v[a/x]\vDash \phi(x,f(x,c)) \iff M'\vDash \forall x \phi(x,f(x,c))$(this is a statement, no assignment needed)

*from our definition of $f^{M'}$ the following claim is also true : $M' \vDash \forall x \forall y: (f(x,c)=y)$ (this is a statement, no assignment needed)

*Therefore also $M' \vDash \forall x \forall y (f(x,c)=y \rightarrow \phi (x,y)) $ because M' satisfies both suffix and prefix. 

*Since the original $M$ satisfied $T$, we get that $M' \vDash T' \Rightarrow M'\vDash \beta$
I'm not sure whether the 5th and 6th bullets are correct and are allowed. I want to verify that this is the correct way of doing this proof.
Thanks
 A: Your proof is along the right lines, but in the last bullet, what you need to show is that $M \models \beta$, because you are trying to show that any model of $T$ is a model of $\beta$.
What you are calling a "preserving extension" is usually called a "conservative extension". In the definition of a conservative extension as it applies to your example, $\beta$ is a formula that does not contain the symbols $f$ and $c$.
Given a model $M$ of $T$, because $M \models \forall x \exists y\phi(x, y)$, you can expand $M$ to a model of $T' = T\cup \lbrace \forall x \forall y (f(x,c)=y \rightarrow \phi)\rbrace$. ("Expanding" a model means giving interpretations to new symbols using the same carrier set: essentially what you are doing in your third bullet.) Conversely, given a model of $T'$, you can obtain a model of $T$ by forgetting about the interpretations of $f$ and $c$. Hence (using the soundness and completeness of first-order logic), if $\beta$ is a sentence that does not contain the symbols $f$ and $c$, $T \models \beta$ iff $T' \models \beta$.
I hope this gives you enough clues to patch up your argument.
