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I am relatively new to the formalism of Mathematical Logic, and don't know how denote the set of the wff logically deducible by a given set of premises $\Sigma$ in a predicate calculus $K.$ I have thought of $\textrm{Theor}(\Sigma),$ as an abbreviation for "theorems of $\Sigma$," but I don't know if it is acceptable.

Motivation: I needed such a notation, because $\Sigma$ is consistent (resp.consistent and complete) iff $\textrm{Theor}(\Sigma)/\tilde{}$ is a proper filter (ultrafilter) in $K^{\star},$ the Lindenbaum algebra of $K.$ In such a way the Lindenbaum's Lemma can be proved just invoking the Ultrafilter Lemma.

So my question is:

There is a standard notation for the set $\{\phi\mid\Sigma\vdash_K\phi\},$ where $K$ is first order predicate calculus and $\Sigma$ is a set of wff in $K$?

I would like to have references to actual usage. Thanks.

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up vote 2 down vote accepted

A formula $\phi$ such that $\Sigma\vdash\phi$ is widely called a "theorem of $\Sigma$". $K$ is usually left implicit if it is one of the many equivalent deduction systems for classical first-order logic, and if there is no need to specify explicitly which language one is working with.

However, I don't think there is any generally used symbolic abbreviation for $\{\phi\mid \Sigma\vdash \phi\}$, the set of theorems of $\Sigma$. None of the references I checked even bother to define one. In most cases where one would use it, it seems to be just as easy to speak of the $\vdash$ relation directly.

On the other hand, if you find yourself needing such a notation, there's nothing wrong with just defining one for yourself. $\operatorname{Theor}(\Sigma)$ or $\operatorname{Thms}(\Sigma)$ would work as well as anything, and shouldn't confuse anyone if only you spend a line introducing the notation before you start using it.

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Dear Henning Makholm thanks a lot for the answer. I needed such a notation, because $\Sigma$ is consistent (and complete) iff $\textrm{Thms}(\Sigma)/\tilde{}$ is a proper filter (ultrafilter) in $K^{\star},$ the Lindenbaum algebra of $K;$ so I can prove Lindenbaum's Lemma just invoking Ultrafilter Lemma. –  Giuseppe Tortorella Jul 29 '12 at 20:22
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