# Finding a finite model

Hello I am having difficulty with this question, I am not even sure what strategy one would go about proving something like this:

Suppose $L$ is a language which includes an infinite list $c_1,c_2,\cdots$ of constant symbols. Let $\Gamma$ be a set of sentences $\Gamma = \{c_i \neq c_j \mid i,j\in N, i < j\}$. Let A be a sentence such that $\Gamma \Rightarrow A$. Prove that $A$ has a finite model.

I am not sure whether I would prove this via a contradiction (i.e., assume $A$ has an infinite model, or if I show a finite model that works or some how assume that we can have an infinite model and then use some sort of compactness to show it can be finite. I am a little all over the place with this question, please any help would be great!

• (Possible) hint: $A$ can only refer to a finite number of constant symbols. – Dave Clarke Apr 24 '14 at 19:49
• Showing that $A$ has an infinite model doesn't give a contradiction: it could have a finite model too. – David Richerby Apr 24 '14 at 19:59

Note that you're looking for a model of $\{A\}$, not of $\Gamma \cup \{A\}$ (which obviously has no finite model since all the constants need to be pairwise distinct).
Proof by contradiction would be “assume $A$ has only infinite models or no model at all”. This isn't likely to lead anywhere.
The theory is $\{A\}$. This is a finite theory, consisting of (one) finite formula. Only a finite number of constants appear in this formula. Let $C(A)$ be the set of constants that appear in $A$. Define $\mathscr{M}: c \mapsto c$ if $c \in C(A)$, $c \mapsto c_1$ if $c \notin C(A)$. Then $\mathscr{M}$ satisfies $A$.