show that if $x$ is not free in $\Psi$ then
$$\phi \rightarrow\Psi\vDash [(\exists x \phi)\rightarrow\Psi]$$ answer: by QR $$\phi \rightarrow\Psi\vdash [(\exists x \phi)\rightarrow\Psi]$$ so theorem soundness $\Rightarrow$$$\phi \rightarrow\Psi\vDash [(\exists x \phi)\rightarrow\Psi]$$ could i tell this way?