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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](http://math.stackexchange.com/questions/82550/how-to-prove-this-models). Note: This is question 1(a) of [this assignment sheet](http://www.student.cs.uwaterloo.ca/~cs245/assign4.pdf).2011-11-16
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    @Henning: I've closed the other one as a duplicate of this one, just because this is the one with the answer in it.2011-11-18
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    @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).2011-11-18

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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?2011-11-16
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    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.2011-11-16
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    Also, don't use `->` to produce an arrow in (La)TeX; it looks horrible. Use `\rightarrow` or its synonym `\to` instead.2011-11-16