I'm told that intuitionistic logic is basically classical logic with "law of the excluded middle removed", and that you can't from $\neg \neg A$ infer $A$.
So, if we consider $A$ a state, $\neg A$ a state, and $\neg \neg A$ a state, does this mean that $\neg \neg \neg A$ is a separable state?
In addition to the two very nice formal answers, let me give an informal proof that $\neg\neg\neg A \Rightarrow \neg A$. Recall that, by definition, $\neg B$ is an abbreviation for $B \Rightarrow \bot$.
Assume $\neg\neg\neg A$. We are to show $\neg A$, that is, $A \Rightarrow \bot$.
So assume $A$, we are to show $\bot$.
Since $A$, we have $\neg\neg A$. But we also have $\neg(\neg\neg A)$. Hence $\bot$.
This proof uses the lemma that $A \Rightarrow \neg\neg A$. This lemma can also be proven informally:
Assume $A$. We are to show $\neg\neg A$, that is, $(\neg A) \Rightarrow \bot$.
So assume $\neg A$. We are to show $\bot$.
Since $\neg A$ (which means $A \Rightarrow \bot$) and since $A$, we have $\bot$.
By the way, at no point did we use the principle of explosion (ex falso quodlibet). Hence our proofs are even valid in minimal logic, where $\bot$ doesn't have any special rules attached to it.
Given $\neg\neg\neg A$:
1. $\neg \neg \neg A$ (premise)
2.1. $A$ (assumption)
2.2.1. $\neg A$ (assumption)
2.2.2. $A$ (reiterate 2)
2.2.3. $\bot$ (contradiction 4 3)
2.3. $\neg\neg A$ ($\neg$intro 2.2.1-2.2.3)
2.4. $\neg\neg\neg A$ (reiterate 1)
2.5. $\bot$ (contradiction 2.3 2.4)
3. $\neg A$ ($\neg$intro 2.1-2.5)
Given $\neg A$:
1. $\neg A$ (premise)
2.1. $\neg \neg A$ (assumption)
2.2. $\neg A$ (reiterate 1)
2.3. $\bot$ (contradiction 2.2 2.1)
3. $\neg\neg\neg A$ ($\neg$intro 2.1-2.3)
Triple negation $\neg \neg \neg P $ is a shorthand for $((P \rightarrow \bot) \rightarrow \bot) \rightarrow \bot$. It turns out however that for any pair of propositions $P$ and $Q$, we have that $((P \rightarrow Q) \rightarrow Q) \rightarrow Q$ simplifies to $P \rightarrow Q$, which lets us simplify triple negation to $P \rightarrow \bot$ which is just the definition of $\neg P$.
To prove this, just note that by modus ponens, we have $P \rightarrow ( (P \rightarrow Q) \rightarrow Q)$, for any P, Q. But we note that the RHS of that is equal to the LHS of $((P \rightarrow Q) \rightarrow Q) \rightarrow Q$ , so we can just compose the two implications to get $P \rightarrow Q$. This holds for any logic with modus ponens and transitive implication.
It is standard to identify $\lnot A$ with $A \to \bot$ in intuitionistic logic. The $\lambda$-term $\lambda f^{((A\to B)\to B) \to B}{\cdot} \lambda a^{A}{\cdot} f(\lambda g^{A \to B}{\cdot} g(a))$ corresponds to a proof of $(((A \to B) \to B) \to B) \to (A \to B)$ under the Curry-Howard Correspondence. Instantiating $B$ to $\bot$ gives you a proof that $\lnot\lnot\lnot A \to \lnot A$.
The proof of the converse proposition $(A \to B) \to (((A \to B) \to B) \to B)$ is witnessed by $\lambda f^{A\to B}{\cdot}\lambda g^{(A \to B) \to B}{\cdot}g(f)$
For a quick proof that $\neg \neg \neg A$ is the same as $\neg A$, first show that $B \to \neg \neg B$.
For the first part, substitute $A$ for $B$ to get $A \to \neg \neg A$. Then using contravariance of $\neg$, we have $\neg \neg \neg A \to \neg A$. The contravariance of $\neg$ is the property that if $C \to D$, then $\neg D \to \neg C$.
For the second part, substitute $\neg A$ for $B$, to get that $\neg A \to \neg \neg \neg A$.
Thus, $\neg A$ and $\neg \neg \neg A$ are logically equivalent.
