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Show that if $\vDash\Phi \to \Psi$ then $\vDash\forall x.\Phi \to \forall x.\Psi$

I started by letting $I,\theta$ be arbitrary

Assume that $I,\theta\vDash\Phi \to \Psi$

Need to show that $I,\theta\vDash\forall x.\Phi \to \forall x.\Psi$

Assume that $I,\theta\vDash\forall x.\Phi$

(for all v $\in \Bbb D$ $I,\theta[x/y]\vDash\forall x.\Phi$)

Need to show that $I,\theta\vDash\forall x.\Psi$

(for all v $\in \Bbb D$ $I,\theta[x/y]\vDash\forall x.\Psi$)

Let v $\in \Bbb D$ be arbitrary,

$I,\theta\vDash\Phi \to \Psi$

This is where I got stuck. I don't know how to introduce the quantifier or x into this bit so I can proceed. Any help would be appreciated.


1 Answer 1

Hint: the basic proof idea will be this. If an interpretation $I$ makes $\forall x\Phi(x)$ true, then any expansion of the interpretation to assign a value to $x$ makes $\Phi(x)$ true. By our initial assumption this expanded interpretation must make $\Psi(x)$ true. But this means that the original interpretation $I$ makes $\forall x\Psi(x)$ true too.

You now just need to fancy up that idea, presenting it in your preferred official idiom.

I said that since all arbitrary $I,\theta\vDash\Phi \to \Psi$ we can say that $I,\theta[x/y]\vDash\Phi \to \Psi$ as well. Can I say this or am I misunderstanding what you mean? –  Brandon Nov 19 '12 at 9:59
You started from : assume that $I,\theta\vDash\Phi \to \Psi$ and : assume that $I,\theta\vDash\forall x.\Phi$. Try by contradiction : if NOT $\vDash\forall x.\Psi$ ; this means that you can find $I$ and an assignment $\alpha$ for which $\Psi$ is false. This contradict $I,\theta\vDash\Phi \to \Psi$ for $I,\theta$ arbitrary. –  Mauro ALLEGRANZA Nov 26 '13 at 16:58

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