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$ \models \phi \to \forall x.\phi$ where $\phi$ is WFF and x not free in $\phi$

Does that render that $ \phi \to \forall x.\phi$ is true in any cases?

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possible duplicate of How to prove this models. Note: This is question 1(a) of this assignment sheet. – Henning Makholm Nov 16 '11 at 23:10
@Henning: I've closed the other one as a duplicate of this one, just because this is the one with the answer in it. – Zev Chonoles Nov 18 '11 at 22:51
@Zev, it's not clear that they are really duplicates -- the OP here seemed to want to know what was being asked of him, whereas in the other question wanted help actually doing it. But in any case the question is now academic (!); the assignment was due yesterday. (Hmm, I see now that the asker was the same. Didn't notice that before). – Henning Makholm Nov 18 '11 at 22:59
up vote 1 down vote accepted

Yes, $\vDash\psi$ means that $\psi$ is logically valid, that is, true in every interpretation of the language of $\psi$.

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so furthermore,to prove $ \models \phi -> \forall x.\phi $ where x not free in $\phi$,I have to figure out that $ \models \phi -> \forall x.\phi $ belongs to WFF? – jason Nov 16 '11 at 1:57
WFF (well-formed formula) just means that $\phi$ has no syntax errors, that is, it is not something like "$)(\land\land\to x(x$". It is usually not something you'll need to spend any explicit effort proving, because you either construct $\phi$ explicitly as something well-formed or get from an assumption that it is. – Henning Makholm Nov 16 '11 at 12:16
Also, don't use -> to produce an arrow in (La)TeX; it looks horrible. Use \rightarrow or its synonym \to instead. – Henning Makholm Nov 16 '11 at 12:17

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