 Proof: We proceed by induction on the number of clauses used to establish \(\Gamma \vdash \theta\). So let \(n\) be a natural number, and assume that the theorem holds for any argument established as deductively valid with fewer than \(n\) steps. And suppose that \(\Gamma \vdash \theta\) was established using exactly \(n\) steps. If the last rule applied was \((=\)I) then \(\theta\) is a sentence in the form \(t=t\), and so \(\theta\) is logically true. A fortiori, \(\Gamma \vDash \theta\). If the last rule applied was (As), then \(\theta\) is a member of \(\Gamma\), and so of course any interpretation that satisfies every member of \(\Gamma\) also satisfies \(\theta\). Suppose the last rule applied is (&amp;I). So \(\theta\) has the form \((\phi \amp \psi)\), and we have \(\Gamma_1 \vdash \phi\) and \(\Gamma_2 \vdash \psi\), with \(\Gamma = \Gamma_1, \Gamma_2\). The induction hypothesis gives us \(\Gamma_1 \vDash \phi\) and \(\Gamma_2 \vDash \psi\). Suppose that \(M\) satisfies every member of \(\Gamma\). Then \(M\) satisfies every member of \(\Gamma_1\), and so \(M\) satisfies \(\phi\). Similarly, \(M\) satisfies every member of \(\Gamma_2\), and so \(M\) satisfies \(\psi\). Thus, by the clause for “\(\amp\)” in the definition of satisfaction, \(M\) satisfies \(\theta\). So \(\Gamma \vDash \theta\). 