Confusion about model theory notes On p.8 of http://www.msri.org/publications/books/Book39/files/marker.pdf, the author writes $\Gamma(\bar{d})$, when $\Gamma$ is, first of all, a set of formulas (not a single one), and it is a formula which has variables, not constants. This doesn't make sense. And what does he mean by $T+\Gamma(\bar{d})$? This would have to mean that we are working in a different language. And why does it imply facts about $\psi_i(\bar{v})$ when we have $\bar{d}$, i.e. a set of constants, rather than variables, when $\bar{v}$ is a set of variables, not a set of constants?
Similarly, in the proof of the "CLAIM," in the sentence that begins "If $\Sigma$ is inconsistent...", how can we go from $\psi_1(\bar{d})$ to $\psi_1(\bar{v})$? One takes constants, the other takes variables.
 A: I think that given David's questions in the comments after the initial answer above, a more informal explanation might help somewhat. You do not necessarily have to think about adding constants to the language as discussed above. 
One way of thinking which might help is the following: in the notation above, lets just think of $\bar d$ as being a tuple of variables. Then $\Gamma (\bar d)$ is simply a collection of formulas where the free variables come from the elements of the tuple $\bar d$. Another way of thinking of the above statement is: 
$$T \vdash \Gamma (\bar d) \rightarrow \phi(\bar d).$$
If you are uncomfortable with the process of changing the language (and I think perhaps this just takes some time to get used to), I hope this helps. 
A: The notation $\Gamma(\bar d)$ means that for every formula $\psi(\bar v)\in\Gamma(\bar v)$ we substitute $\bar d$ for $\bar v$. The set $T+\Gamma(\bar d)$ is just the union $T\cup\Gamma(\bar d)$.
The interchanging of variables $\bar v$ and constants $\bar d$ in the proof comes from the fact that the constants $\bar d$ are new constants that do not appear in $T$. Thus, for example, if $T\models \phi(\bar d)$, then also $T\models\forall \bar v (\phi(\bar v))$.
