 Proof: The proof of completeness is rather complex. We only sketch it here. Let \(\Gamma\) be a consistent set of sentences of \(\LKe\). Again, we assume for simplicity that the set \(K\) of non-logical terminology is either finite or countably infinite (although the theorem holds even if \(K\) is uncountable). The task at hand is to find an interpretation \(M\) such that \(M\) satisfies every member of \(\Gamma\). Consider the language obtained from \(\LKe\) by adding a denumerably infinite stock of new individual constants \(c_0, c_1,\ldots\) We stipulate that the constants, \(c_0, c_1,\ldots\), are all different from each other and none of them occur in \(K\). One interesting feature of this construction, due to Leon Henkin, is that we build an interpretation of the language from the language itself, using some of the constants as members of the domain of discourse. Let \(\theta_0 (x), \theta_1 (x),\ldots\) be an enumeration of the formulas of the expanded language with at most one free variable, so that each formula with at most one free variable occurs in the list eventually. Define a sequence \(\Gamma_0, \Gamma_1,\ldots\) of sets of sentences (of the expanded language) by recursion as follows: \(\Gamma_0 = \Gamma\); and \(\Gamma_{n+1} = \Gamma_n,(\exists x\theta_n \rightarrow \theta_{n}(x|c_i))\), where \(c_i\) is the first constant in the above list that does not occur in \(\theta_n\) or in any member of \(\Gamma_n\). The underlying idea here is that if \(\exists x\theta_n\)is true, then \(c_i\) is to be one such \(x\). Let \(\Gamma\) be the union of the sets \(\Gamma_n\). 