1
$\begingroup$

I was told that:

$$ Th(\Sigma) := \{ \sigma : \Sigma \vdash \sigma \} $$

is an L-theory (i.e. its closed under provability i.e. $if T \vdash \sigma \implies \sigma \in T$). I feel it should be a trivial proof but its escaping me right now. Why is that true?


My thoughts:

We want to show (WTS) that $T \vdash \sigma \implies \sigma \in T$. So assume $T \vdash \sigma$ holds, does $\sigma \in T$ hold? Well if the hypothesis holds then there is a finite sequence s.t. $p=p_1...p_n$ s.t.

  1. $p_i$ is an axiom
  2. $p_i \in T$
  3. $p_i$ is an Inference Rule

what I want to conclude is that $\Sigma \vdash p_i$ so something in that list must mean that. I assume it must be something about step 2 but it escapes me...is this suppose to be easy or perhaps its not as easy as I thought?

$\endgroup$
3
$\begingroup$

Oh I see. Its because we defined $T = Th(\Sigma) = \{ \sigma : \Sigma \vdash \sigma \}$ so by definition everything in $T$ is provable in $\Sigma$. So step 2 implies $\Sigma \vdash \sigma$. So it automatically becomes a proof in $\Sigma$. So $\Sigma \vdash p$ which means $\sigma \in T$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.