Proof: $Th(T)=Th(Th(T))$
We need to show that:
- $Th(T)\subseteq Th(Th(T))$ (1)
- $Th(Th(T))\subseteq Th(T)$ (2)
(1) is easily achieved by apply inflationaryness property to $Th(T)$.
To prove (2), assume that $A\in Th(Th(T))$, we need to show that $A\in Th(T)$. Indeed:
$A\in Th(Th(T))\Rightarrow Th(T)\vdash A\Rightarrow$ either:
- $A\in Th(T)$, or
- $A$ is an axiom $\Rightarrow A\in Th(T)$, or
- there is a derivation $A_1,\ldots,A_n$ of $Th(T)$ s.t. $A_n=A$. We call it derivation (a)
In this derivation, either $A_i\in Th(T)$, $A _i$ is an axiom, or $A_i$ is a direct consequence of earlier elements under some inference rule.
If $A_i\in Th(T)\Rightarrow T\vdash A$, then there is a derivation (b): $B_1,\ldots,B_m$ of $T$ s.t. $B_m=A_i$ and: either $B_j\in T$, or $B_j$ is an axiom, or $B_j$ is a direct consequence of earlier elements under some inference rule.We can replace $A_i$ in derivation (a) by derivation (b) without destroying (a)'s correctness.
By doing this replacement such $A_i$, we will have another equivalent derivation (a') for A such that each element is either in $T$ or an axiom or a direct consequence of earlier elements under some inference rule.
That means $T\vdash A$, hence $A\in Th(T)$
So, in all cases, from $A\in Th(Th(T))\Rightarrow A\in Th(T)$. (2) is proved.
The idempotency property is proved.