Enderton's logic book about completeness theorem In page 141 of A Mathematical Introduction to Logic, Enderton simply writes,

STEP 6: Restrict the structure $\mathfrak{A}/E$ to the original language. This
  restriction of $\mathfrak{A}/E$ satisfies every member of $\Gamma$ with $h \circ s$.

where $\mathcal{L'}=\mathcal{L}\cup C$($C$ is the set of new countable constants), and $\mathfrak{A}/E$ is a quotient structure for $\mathfrak{A}$.
I don't get what STEP 6 means and how to prove. Could you give some idea for this?
 A: The step is already present in the part of the proof regarding the language without the equality symbol [pag 140, first paragraph].
In order to manufacture the structure $\mathfrak A$, we have to [page 135 : Step 1] :

Expand the language by adding a countably infinite set of new constant symbols. 

The domain of the structure $\mathfrak A$ satisfying $\Delta \supseteq \Gamma$ is the set of all terms of the new language [page 137].
But we have to remember taht a structure $\mathfrak A$ is not only its universe (or domain) $|\mathfrak A|$ [see page 80] :

Formally, a structure $\mathfrak A$ for our given first-order language is a function whose domain is the set of parameters and such that :

[...]



    
*$\mathfrak A$ assigns to each constant symbol $c$ a member $c^{\mathfrak A}$ of the universe $|\mathfrak A|$.
    
  

Conclusion : having expanded the original language with the new individual constants, the structure $\mathfrak A$ assign also a reference to each of them.
Those "references" are not needed in a structure for $\Gamma$, because the formulae in it are relative to the original language, and thus they have no occurrence of the added constants.
So we have to "throw away" from $\mathfrak A$ those objcets and what we get is the restriction of $\mathfrak A$ to the original language; the restricted structure is a structure that satisfies every member of $\Gamma$.
