 We need to show that \(\Gamma\vDash\theta\). By the induction hypothesis, we have that \(\Gamma_1\vDash\exists v\phi\) and \(\Gamma_2, \phi(v|t)\vDash\theta\). Let \(M\) be an interpretation such that \(M\) makes every member of \(\Gamma\) true. So, \(M\) makes every member of \(\Gamma_1\) and \(\Gamma_2\) true. Then \(M,s\vDash\exists v\phi\) for all variable assignments \(s\), so there is an \(s'\) such that \(M,s'\vDash\phi\). Let \(M'\) differ from \(M\) only in that \(I_{M'}(t)=s'(v)\). Then, \(M',s'\vDash\phi(v|t)\) and \(M',s'\vDash\Gamma_2\) since \(t\) does not occur in \(\phi\) or \(\Gamma_2\). So, \(M',s'\vDash\theta\). Since \(t\) does not occur in \(\theta\) and \(M'\) differs from \(M\) only with respect to \(I_{M'}(t)\), \(M,s'\vDash\theta\). Since \(\theta\) is a sentence, \(s'\) doesn't matter, so \(M\vDash\theta\) as desired. Notice the role of the restrictions on \((\exists\)E) here. The other cases are about as straightforward. 