 Proof: Suppose \(\Gamma_1 \vdash \psi\) and \(\Gamma_2, \psi \vdash \theta\). We proceed by induction on the number of rules used to establish \(\Gamma_2, \psi \vdash \theta\). Suppose that \(n\) is a natural number, and that the theorem holds for any argument that was derived using fewer than \(n\) rules. Suppose that \(\Gamma_2, \psi \vdash \theta\) was derived using exactly \(n\) rules. If the last rule used was \((=\)I), then \(\Gamma_1, \Gamma_2 \vdash \theta\) is also an instance of \((=\)I). If \(\Gamma_2, \psi \vdash \theta\) is an instance of (As), then either \(\theta\) is \(\psi\), or \(\theta\) is a member of \(\Gamma_2\). In the former case, we have \(\Gamma_1 \vdash \theta\) by supposition, and get \(\Gamma_1, \Gamma_2 \vdash \theta\) by Weakening (Theorem 8). In the latter case, \(\Gamma_1, \Gamma_2 \vdash \theta\) is itself an instance of (As). Suppose that \(\Gamma_2, \psi \vdash \theta\) was obtained using (&amp;E). Then we have \(\Gamma_2, \psi \vdash(\theta \amp \phi)\). The induction hypothesis gives us \(\Gamma_1, \Gamma_2 \vdash(\theta \amp \phi)\), and (&amp;E) produces \(\Gamma_1, \Gamma_2 \vdash \theta\). The remaining cases are similar. 