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?


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 $\sigma \in T$.


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