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**HTale** $\displaystyle \text{Cn}(\text{Cn}(A)) = \text{Cn}(A)$ where, $\displaystyle Cn(A) = \{\beta : A \vdash \beta \}$

Secondly, pick a $\displaystyle \gamma \in \text{Cn}(\text{Cn}(A)) $. Then this is such that for some $\displaystyle \beta$, with $\displaystyle A \vdash \beta$, we have that $\displaystyle \beta \vdash \gamma$. But, given the transitivity property, $\displaystyle A \vdash \gamma$, to conclude that $\displaystyle \gamma \in \text{Cn}(A)$. Thus, $\displaystyle \text{Cn}(\text{Cn}(A)) \subseteq \text{Cn}(A)$.

Combining the two, we conclude that $\displaystyle \text{Cn}(\text{Cn}(A)) = \text{Cn}(A)$.