Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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 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.

share|cite|improve this question
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.

share|cite|improve this answer
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 Jul 29 '12 at 20:22

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.