# Why is $Th(\Sigma) := \{ \sigma : \Sigma \vdash \sigma \}$ an L-theory?

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 $$\sigma \in T$$.