0
$\begingroup$

Here is an exercise (from some lecture notes about mathematical logic):

Prove that every maximally consistent set of formulas is a theory.

Here is the solution given in the lecture notes.

Assume that Γ is maximally consistent. Suppose that Γ ⊢α. Then, by the consistency of Γ, ¬α is not derivable from Γ. This entails that ¬α ∉ Γ, otherwise we would get that Γ ⊢ ¬α. By the maximality of Γ, it follows that α ∈ Γ.

I don't understand exactly why this solution constitutes a proof of the fact that we are dealing with a theory.

Could you help me please?

Thank you very much

Fish

  • 3
    What is the definition of "theory" ? Closed under derivability ? – 2017-02-27
  • 1
    What's the definition of a theory that was given in the notes ? In most definitions I have seen, a theory is simply a set of sentences... – 2017-02-27
  • 0
    If *theory* means "closed under derivability", to show that $\Gamma$ is a *theory* means to prove that : if $\Gamma \vdash \alpha$, then $\alpha \in \Gamma$. – 2017-02-27
  • 0
    In the notes a theory is defined as a deductively closed set o formulas. (I don't see which strange symbol you're talking about). – 2017-02-27
  • 0
    See [van Dalen](https://books.google.it/books?id=u0wlXPHATDcC&pg=PA98), page 98 : "A *theory* $T$ is a collection of sentences with the property $T \vdash \varphi \Rightarrow \varphi \in T$ (a theory is *closed under derivability*)." – 2017-02-27
  • 0
    Thank you Mauro. I know it's a dumb question, but I just wanted to be certain. – 2017-02-27

1 Answers 1

2

Note that there are two definitions of "theory" - a deductively closed set of sentences (which is what's being used here), or any set of sentences.

Since you care about deductively closed sets of sentences, let me rephrase the theorem:

Any maximal consistent set of sentences is deductively closed.

The idea behind why this is true is that - if $\Gamma$ is maximal consistent, and $\Gamma\vdash\alpha$ - then $\alpha$ had better be in $\Gamma$, or else $\Gamma$ wouldn't be maximal. Specifically, we show that $\Gamma\cup\{\alpha\}$ is consistent - this means $\Gamma\cup\{\alpha\}=\Gamma$ (since $\Gamma$ is maximal consistent), i.e., $\alpha\in\Gamma$.

So how do we show that $\Gamma\cup\{\alpha\}$ is consistent? Assume for contradiction that it isn't. Then - by definition - $\Gamma\vdash \neg\alpha$. But then $\Gamma$ is inconsistent, since $\Gamma\vdash\alpha$ and $\Gamma\vdash\neg\alpha$. And we assumed $\Gamma$ was consistent, so this is a contradiction.

This is exactly what the proof you've described is doing, and the conclusion - that if $\alpha$ is derivable from $\Gamma$, then $\alpha$ is in $\Gamma$ - is exactly the statement "$\Gamma$ is deductively closed", or in other words "$\Gamma$ is a theory".