Prove or disprove: if $\Sigma$ is special then $\left\{ \lnot\alpha\,:\,\alpha\in\Sigma\right\} $ is special Practicing for exam in Logic and trying to solve the following question:

Let $\Sigma\subseteq WFF_{\left\{ \lnot,\to\right\} }$ be set of propositions. $\Sigma$ is called special if for every $\alpha,\beta\in WFF_{\left\{ \lnot,\to\right\} }$ we get $\Sigma\vdash\alpha\to\beta$ or $\Sigma\vdash\beta\to\left(\lnot\alpha\right)$. Prove or disprove: if $\Sigma$ is special then $\left\{ \lnot\alpha\,:\,\alpha\in\Sigma\right\} $ is special.

I'm suspecting that this statment is not true. I choose $\Sigma$ to be $WFF_{\left\{ \lnot,\to\right\} }$ and I proved that $\Sigma$ is special because you can show a proof sequence (because $\left(\alpha\to\beta\right)\in\Sigma$). But how do I disprove $\left\{ \lnot\alpha\,:\,\alpha\in\Sigma\right\} $ to be special? I know the theorem that for every $\Sigma\subseteq WFF_{\left\{ \lnot,\to\right\} }$ we get $\Sigma\vDash\alpha$ iff $\Sigma\vdash\alpha$. I think it might help here.
 A: By $WFF_{\{\neg, \rightarrow\}}$ I assume all of your formulas have only $\neg$ and $\rightarrow$ as connectives? This will make things a bit less clear...
I believe you can find a special set $\Sigma$ such that $\neg\Sigma=\{\neg\alpha : \alpha\in\Sigma\}$ is not special: consider the propositional variables $p$ and $q$ and define
$$\Sigma=\{\neg p, \neg(p\rightarrow\neg q)\},$$
but notice that $\neg(p\rightarrow \neg q)$ is equivalent to $p\wedge q$; this means that the logical closure of $\Sigma$ contains all (well-formed) formulas, and so $\Sigma$ is special.
(As you noticed, you can test this by using truth tables: if both $\neg p$ and $\neg(p\rightarrow\neg q)$ are true, $p$ and $p\rightarrow \neg q$ are false; but if $p\rightarrow \neg q$ is false, $p$ is true and $\neg q$ is false, what would lead to a contradiction. This means $\Sigma$ is contradictory, and in particular can derive any formula, meaning it is special.)
Now, $\neg\Sigma=\{p, p\rightarrow\neg q\}$ (up to equivalences), and notice that $p\rightarrow\neg q$ is equivalent to $\neg p\vee \neg q$. We can easily find formulas $\alpha$ and $\beta$ such that neither $\neg\Sigma\vdash \alpha\rightarrow \beta$ nor $\neg\Sigma\vdash\beta\rightarrow\neg\alpha$: take, for example, new propositional variables $r$ and $s$, $\alpha=r$ and $\beta=s$.

*

*If $p$ and $r$ are true but $q$ and $s$ are false, $\neg\Sigma$ is satisfied but not $r\rightarrow s$, meaning $\neg\Sigma$ does not prove $\alpha\rightarrow\beta$.

*If $p$, $r$ and $s$ are true but $q$ is false, $\neg\Sigma$ is satisfied but not $s\rightarrow\neg r$, meaning $\neg\Sigma$ does not prove $\beta\rightarrow\neg\alpha$.

