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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!

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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
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@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

1 Answer 1

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.

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