 Proof: By Theorem 1 and Lemma 3, if \(\alpha\) contains a left parenthesis, then the right parenthesis that matches the leftmost left parenthesis in \(\alpha \beta\) comes at the end of \(\alpha \beta\), and so the matching right parenthesis is in \(\beta\). So, \(\alpha\) has more left parentheses than right parentheses. By Theorem \(1, \alpha\) is not a formula. So now suppose that \(\alpha\) does not contain any left parentheses. By Lemma \(2, \alpha \beta\) consists of a string of zero or more unary markers followed by either an atomic formula or a formula produced using a binary connective, via one of clauses (3)–(5). If the latter formula was produced via one of clauses (3)–(5), then it begins with a left parenthesis. Since \(\alpha\) does not contain any parentheses, it must be a string of unary markers. But then \(\alpha\) does not contain any atomic formulas, and so by Lemma \(4, \alpha\) is not a formula. The only case left is where \(\alpha \beta\) consists of a string of unary markers followed by an atomic formula, either in the form \(t_1 =t_2\) or \(Pt_1 \ldots t_n\). Again, if \(\alpha\) just consisted of unary markers, it would not be a formula, and so \(\alpha\) must consist of the unary markers that start \(\alpha \beta\), followed by either \(t_1\) by itself, \(t_1 =\) by itself, or the predicate letter \(P\), and perhaps some (but not all) of the terms \(t_1, \ldots,t_n\). In the first two cases, \(\alpha\) does not contain an atomic formula, by the policy that the categories do not overlap. Since \(P\) is an \(n\)-place predicate letter, by the policy that the predicate letters are distinct, \(P\) is not an \(m\)-place predicate letter for any \(m \ne n\). So the part of \(\alpha\) that consists of \(P\) followed by the terms is not an atomic formula. In all of these cases, then, \(\alpha\) does not contain an atomic formula. By Lemma \(4, \alpha\) is not a formula.