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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.

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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 $T\models\forall \bar v (\phi(\bar v))$.

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  • $\begingroup$ So you're saying that for each $\bar{v}$, we can apply the new language (i.e. including the $d_i$'s) to this structure by interpreting $d_1$ as $v_1$, etc? I.e. we make a separate application of the language to our structure for each $\bar{v}$. $\endgroup$
    – David Corwin
    Commented Jul 26, 2010 at 23:04
  • $\begingroup$ So you're saying that for each $\bar{v}$, we can apply the new language (i.e. including the $d_i$'s) to this structure by interpreting $d_1$ as $v_1$, etc? I.e. we make a separate application of the language to our structure for each $\bar{v}$. $\endgroup$
    – David Corwin
    Commented Jul 26, 2010 at 23:05
  • $\begingroup$ @Davidac897: Not quite. Remember that an interpretation of a constant in a structure must be an element of that structure, not a variable symbol such as v_1. I see that in the first paragraph of the proof of the CLAIM, Marker is using a standard "abuse of notation" in model theory where the same symbol d_1 is being used both for a new constant symbol added to the language of T, and for the interpretation of this constant in a particular structure. Some other books (e.g. Chang and Keisler) have notation to carefully distinguish these two things, but it looks like Marker's doesn't. $\endgroup$
    – John Goodrick
    Commented Jul 27, 2010 at 2:42
  • $\begingroup$ A "justification" of this way of speaking is that we can always regard an element of the domain of a structure as a constant symbol that that denotes itself. Hodges mentions this explicitly in his "Model Theory". $\endgroup$
    – Sidney Raffer
    Commented Jul 27, 2010 at 6:06
  • $\begingroup$ @John Goodrick: Right, but what I mean is, for a given value of $v_1$, we interpret $d_1$ to be $v_1$? $\endgroup$
    – David Corwin
    Commented Jul 27, 2010 at 8:32
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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.

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  • $\begingroup$ Right, this makes more sense. $\endgroup$
    – David Corwin
    Commented Jul 28, 2010 at 12:48
  • $\begingroup$ +1. And if you look in many standard introductory logic texts (e.g. Enderton's), you can find theorems stating that these two approaches are equivalent (I think this fact is usually called "generalization over constants"). $\endgroup$
    – John Goodrick
    Commented Aug 3, 2010 at 18:32
  • $\begingroup$ Thanks John, I had not actually ever heard the name for such a thing. $\endgroup$
    – James Freitag
    Commented Aug 4, 2010 at 7:34

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