I need to prove that every theory eliminates quantifiers in an appropriate definitional expansion. For this, consider: let $T$ be a theory in language $L$. Consider the following expansion of the language: for each $L$- formula $\varphi(\bar x)$ add a relation $R_\varphi$, whose arity is the length of $\bar x$. Consider the theory $T'$ obtained by adding, for each $L$-formula $\varphi(\bar x)$, the axiom $\forall \bar x (\varphi (\bar x)\leftrightarrow R_\varphi(\bar x))$ to $T$. Prove that:
Each model of $T$ has a unique expansion to a model of $T'$.
If $\varphi$ is an $L$-sentence and $T'\models \varphi$ then $T\models \varphi$.
For every $\varphi (\bar x ) \in L'$ there is $\psi(\bar x) \in L$ such that $T' \models \varphi (\bar x) \leftrightarrow \psi(\bar x)$.
$T'$ has quantifier elimination.
So, for the first item, I have to do two things, one is to show that there is an expansion and the other that it is unique. I guess that to prove that it is unique I need to take 2 "different" expansions and show that they are the same. I don't really know how to properly start writing this, so I would appreciate your input.
For the second one... it seems "evident", but I don't know how to justify it. Assume $T' \models \varphi$, so $T' \models \forall \bar x (\varphi (\bar x)\leftrightarrow R_\varphi(\bar x))$ (can i say that?), I don't really know how to go from here.
I'm really at lost about how to continue, and I would really appreciate your help explaining me how to go about this.