I'm wodering if this statement is provable in logic $ \lnot \alpha \to \lnot \lnot \lnot \alpha ) $ I've encountered this statement in my final exam
$$ \lnot  \alpha \to \lnot \lnot \lnot  \alpha ) $$
there was no open parenthesis and from what I know this is invalid (not a well-formed formula) so I just put NWFF in the answer sheet.
I'm I missing something? Can it be proved in natural deduction or sequent calculus ?   
EDIT : to those who says that this is probably a typo on the exam , no it's not a typo because I've seen the same statement in the exam of the last year !  
 A: It's probably just a typo in the exam.
Anyway note that it would never be a well-formed formula (wff) anyway as long as $\alpha$ plays the role of a metavariable. If you still want a demonstration that any instance of this formula schema is not a wff, it goes like this:
Definition (well-formed formula)


*

*$\alpha$ is a wff if $\alpha$ is atomic

*$(\neg\alpha)$ is a wff if $\alpha$ is a wff

*$(\alpha \rightarrow \beta)$ is a wff if $\alpha$ and $\beta$ are wff


Therefore, by making explict the omitted parenthesis in the OP's formula we have
$$((¬α)→(¬(¬(¬α))))\color{red}{)}$$
Where any instance of it is not a wff, since it would have one additional occurrence of parenthesis - by the way, it's always a good exercise to show the number of parenthesis of a wff is always even (prove it)

The proof schema
If there's any doubt about how to prove any instance of the (well-formed) formula schema:
$$¬α→¬¬¬α$$
(where I omit the parenthesis) in natural deduction it goes like this:

  
*
  
*$\neg\alpha$, Hypothesis 
  
  
  

    
*$\neg(\neg\alpha)$, Hypothesis 
    
*$\neg\alpha \wedge \neg(\neg\alpha)$, 1,2, $\wedge$-Introduction


  
*$\neg(\neg(\neg\alpha))$, 1-3, $\neg$-Introduction
  
*$\neg\alpha \rightarrow \neg(\neg(\neg\alpha))$, 1-4, $\rightarrow$-Introduction

Note that this is not a proof, but only a proof schema.
