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Σ is "proofwise stronger" than a set Γ if {$\alpha$: Σ ⊢ $\alpha$} $\supseteq$ {$\alpha$ : Γ ⊢ $\alpha$ }. Show that for every maximally consistent set of propositions Σ, for every set Γ, either Σ is proofwise stronger than Γ or Σ $\cup$Γ is not consistent, or both (I am confused about "or both").


I was able to prove either Σ is proofwise stronger than Γ or Σ $\cup$Γ is not consistent. However, the "or both" part confuses me. I handed in my proof without showing that they could be "both" and for some reason the marker give me full marks. But then the professor explicitly said we need to prove "or both" in class and on the assignment. Perhaps "or both" is some trivial case? Could someone clarify what is going on here?

My proof (sketch):

If Σ is maximally consistent, then $\forall \alpha,$ Σ ⊢ $\alpha$ or Σ ⊢ $\neg\alpha$.

Suppose Σ is not proofwise stronger than Γ then $\exists\beta$ such that Γ⊢$\beta$, but Σ $\nvdash$ $\beta$. if Σ ⊢ $\neg\beta$ then Σ $\cup$ Γ is not consistent (by monotonicity and the definition of consistency).

but if Σ $\nvdash$ $\neg\beta$ then Σ $\nvdash$ $\neg\beta$ and Σ $\nvdash$ $\beta$. This fails the definition of maximally consistent. Therefore Σ is proofwise stronger.

how to do or both??

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@user6312 I am pretty sure it was "or both". Anyways if you are confused as well then I am going to the professor tomorrow to clarify. – Mark Jul 13 '11 at 6:31
@user6312 Btw for my logic course there are no course notes or textbook, there are only lectures. Do you know a good logic book for this kind of problem? – Mark Jul 13 '11 at 6:37
Maybe I am too easily confused this late. What is the local definition of maximally consistent? There are two choices: (i) is consistent but no strictly stronger theory is or (ii) there is no strictly stronger consistent theory. Definition (ii) would allow $\Sigma$ to be inconsistent. You need to look up the formal definition that was used. If it was (ii), then "or both" is right. Unfortunately in logic there is not the same universality of definitions that one finds in many other fields. That's why I referred to "local" definition, there is variation. – André Nicolas Jul 13 '11 at 6:42
@user6312 It's like I mentioned there are only lectures and the professor is good that he doesn't even bring notes to class. So the course material is being taught in unconventional order. It feels like the course is going in all directions. If you know any good reference that explains the difference between local and universal it would be great! – Mark Jul 13 '11 at 6:46
up vote 4 down vote accepted

The answer hinges on the definition that the course uses for maximally consistent.

On no very good evidence, apart from the fact that you have written down the sentence in your argument, it is possible that "maximally consistent" was defined in your course as follows:

"$\Sigma$ is maximally consistent if $\forall \alpha$, $\Sigma \vdash \alpha$ or $\Sigma \vdash \neg\alpha$."

In that case, any inconsistent $\Sigma$ is automatically maximally consistent. Then the "or both" can certainly hold. If $\Sigma$ is inconsistent, then it is also automatically "proofwise stronger" than $\Gamma$.

It is possible that the definition of "maximally consistent" was different from the guess above, but still allowed for the possibility of inconsistency.

If however, the definition of maximally consistent given in the course is (the more sensible to me) "consistent, but such that no proper extension is consistent" then the "or both" of your question is not possible.

So I am afraid there is no simple answer. You have to check carefully the actual definitions that are used in your course. There is unfortunately no complete uniformity of terminology.

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+1 you are right I will need to get some clarification from the Professor. – Mark Jul 13 '11 at 7:30
It's possible that too much is being read into the question. Maybe the instructor simply wanted the students to prove "at least one of the following holds" instead of the stronger "exactly one of the following holds" (for whatever reason -- exercises often aren't tight results), and didn't use the best possible language. – user83827 Jul 13 '11 at 11:32

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