I'm wodering if this statement is provable in logic $ \lnot \alpha \to \lnot \lnot \lnot \alpha ) $

Question

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 !

Answer 1

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)

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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:

1. $\neg\alpha$, Hypothesis

2. $\neg(\neg\alpha)$, Hypothesis

3. $\neg\alpha \wedge \neg(\neg\alpha)$, 1,2, $\wedge$-Introduction

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

5. $\neg\alpha \rightarrow \neg(\neg(\neg\alpha))$, 1-4, $\rightarrow$-Introduction

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