 Proof: Although this theorem holds in general, we assume here that the set \(K\) of non-logical terminology is either finite or denumerably infinite (i.e., the size of the natural numbers, usually called \(\aleph_0)\). It follows that there is an enumeration \(\theta_0, \theta_1,\ldots\) of the sentences of \(\LKe\), such that every sentence of \(\LKe\) eventually occurs in the list. Define a sequence of sets of sentences, by recursion, as follows: \(\Gamma_0\) is \(\Gamma\); for each natural number \(n\), if \(\Gamma_n, \theta_n\) is consistent, then let \(\Gamma_{n+1} = \Gamma_n, \theta_n\). Otherwise, let \(\Gamma_{n+1} = \Gamma_n\). Let \(\Gamma'\) be the union of all of the sets \(\Gamma_n\). Intuitively, the idea is to go through the sentences of \(\LKe\), throwing each one into \(\Gamma'\) if doing so produces a consistent set. Notice that each \(\Gamma_n\) is consistent. Suppose that \(\Gamma'\) is inconsistent. Then there is a sentence \(\theta\) such that \(\Gamma'\vdash \theta\) and \(\Gamma'\vdash \neg \theta\). By Theorem 9 and Weakening (Theorem 8), there is finite subset \(\Gamma''\) of \(\Gamma'\) such that \(\Gamma''\vdash \theta\) and \(\Gamma''\vdash \neg \theta\). Because \(\Gamma''\) is finite, there is a natural number \(n\) such that every member of \(\Gamma''\) is in \(\Gamma_n\). So, by Weakening again, \(\Gamma_n \vdash \theta\) and \(\Gamma_n \vdash \neg \theta\). So \(\Gamma_n\) is inconsistent, which contradicts the construction. So \(\Gamma'\) is consistent. Now suppose that a sentence \(\theta\) is not in \(\Gamma'\). We have to show that \(\Gamma', \theta\) is inconsistent. The sentence \(\theta\) must occur in the aforementioned list of sentences; say that \(\theta\) is \(\theta_m\). Since \(\theta_m\) is not in \(\Gamma'\), then it is not in \(\Gamma_{m+1}\). This happens only if \(\Gamma_m, \theta_m\) is inconsistent. So a pair of contradictory opposites can be deduced from \(\Gamma_m,\theta_m\). By Weakening, a pair of contradictory opposites can be deduced from \(\Gamma', \theta_m\). So \(\Gamma', \theta_m\) is inconsistent. Thus, \(\Gamma'\) is maximally consistent.