 Proof: Add a collection of new constants \(\{c_{\alpha} | \alpha \lt \kappa \}\), of size \(\kappa\), to the language, so that if \(c\) is a constant in \(K\), then \(c_{\alpha}\) is different from \(c\), and if \(\alpha \lt \beta \lt \kappa\), then \(c_{\alpha}\) is a different constant than \(c_{\beta}\). Consider the set of sentences \(\Gamma'\) consisting of \(\Gamma\) together with the set \(\{\neg c_{\alpha}=c_{\beta} | \alpha \ne \beta \}\). That is, \(\Gamma'\) consists of \(\Gamma\) together with statements to the effect that any two different new constants denote different objects. Let \(\Gamma''\) be any finite subset of \(\Gamma'\), and let \(m\) be the number of new constants that occur in \(\Gamma''\). Then expand the interpretation \(M_m\) to an interpretation \(M_m'\) of the new language, by interpreting each of the new constants in \(\Gamma''\) as a different member of the domain \(d_m\). By hypothesis, there are enough members of \(d_m\) to do this. One can interpret the other new constants at will. So \(M_m\) is a restriction of \(M_m'\). By hypothesis (and Theorem 15), \(M'_m\) satisfies every member of \(\Gamma\). Also \(M'_m\) satisfies the members of \(\{\neg c_{\alpha}=c_{\beta} | \alpha \ne \beta \}\) that are in \(\Gamma''\). So \(M'_m\) satisfies every member of \(\Gamma''\). By compactness, there is an interpretation \(M = \langle d,I\rangle\) such that \(M\) satisfies every member of \(\Gamma'\). Since \(\Gamma'\) contains every member of \(\{\neg c_{\alpha}=c_{\beta} | \alpha \ne \beta \}\), the domain \(d\) of \(M\) must be of size at least \(\kappa\), since each of the new constants must have a different denotation. By Theorem 15, the restriction of \(M\) to the original language \(\LKe\) satisfies every member of \(\Gamma\).