Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.