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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need to show that this formula $$(\forall x(A \to B) \to (\forall x A\to \forall x B))$$ is true for all interpretation. Could you help me please?

Thank you!

share|cite|improve this question
What rules of inference do you have for your system? What axioms do you have? (stock reply often used on the philosophy forums, but it works rather well). – Doug Spoonwood Sep 15 '11 at 23:17
@Doug: If fara worded the question correctly, what matters is the definition of interpretation in a model being used, not the axioms and rules of inference. – Brian M. Scott Sep 15 '11 at 23:51
Your statement is of the form $P \Rightarrow (Q \Rightarrow R)$. Try a proof by contradiction, i.e. assume that $P,Q$ holds, but $R$ doesn't hold, and show that you get a contradiction. – TMM Sep 15 '11 at 23:52
@Brian If he rigorously shows it, how he shows it depends on the rules of inference and axioms he has. If you have a formal proof of a formula, and the system comes as sound, then you do have the formula as true for all interpretations by soundness. – Doug Spoonwood Sep 16 '11 at 1:10
@Doug: The question as phrased is asking for a model-theoretic argument. It’s a question about semantics, not syntax. – Brian M. Scott Sep 16 '11 at 5:48

If $\forall x(Ax \rightarrow Bx)$ is false, formula is true by (Tarski's inductive) definition of truth.

If $\forall x(Ax \rightarrow Bx)$ is true, then by d.o.t., for any element $d \in D$ (where $D$ is your model) it's true that $Ad \rightarrow Bd$.

The consequent says $\forall x Ax \rightarrow \forall x Bx$. By d.o.t., if $\forall x Ax$ is false, the consequent is true, and thus the formula is true.

If however $\forall x Ax$ is true, by d.o.t. we have that for all $d \in D$, $Ad$. So by combining this with $Ad \rightarrow Bd$, by d.o.t. we have that for all $d \in D$, $Bd$, and this is defined to be same as $\forall x Bx$.

Since regardless of model (interpretation) the formula holds, it holds for all interpretations.

share|cite|improve this answer
This is not how you prove for any. To do that you have to pick arbitrary interpretation and using it prove sentence. – Trismegistos Dec 7 '14 at 17:19
That's what is desribed (implicitly), by not referring to any specific interpretation, it was proven for any (arbitrary) interpretation, and thus for all. – Luka Mikec Dec 8 '14 at 15:36

Your Answer


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.