Negation of a Formula is Provable without Including the Formula as an Assumption

Question

The following lemma states that if we can prove negation of a Well Formed Formula (WFF) $\alpha$ by assuming the formula itself, then we can do it without such an assumption.

Lemma. Let $\Sigma$ be a set of WFFs and $\alpha$ a WFF. If $\Sigma \cup \{\alpha\} \vdash (\neg \alpha)$, then $\Sigma \vdash (\neg \alpha)$.

Proof. Suppose that we have shown that $\vdash ((\alpha \to (\neg \alpha)) \to (\neg \alpha))$. By the deduction theorem, $\Sigma \cup \{\alpha\} \vdash (\neg \alpha)$ implies that $\Sigma \vdash (\alpha \to (\neg \alpha))$. By monotonicity, we also have $\Sigma \vdash ((\alpha \to (\neg \alpha)) \to (\neg \alpha))$. Now, applying Modus Ponens (MP), we can conclude that $\Sigma \vdash (\neg \alpha)$.

As you can see, the proof relies on showing that $\vdash ((\alpha \to (\neg \alpha)) \to (\neg \alpha))$. Here is a list of tools that I can use for proving this.

Hilbert Axioms. Any WFF of the the following forms is considered as an axiom.

• $(\alpha \to (\beta \to \alpha))$.
• $\big((\alpha \to (\beta \to \gamma)) \to ((\alpha \to \beta) \to (\alpha \to \gamma))\big)$.
• $\big(((\neg \beta) \to (\neg \alpha)) \to (\alpha \to \beta)\big)$.

Also, by using Hilbert axioms, I have proved the followings.

• $\vdash (\alpha \to \alpha)$.
• $\vdash \big((\neg(\neg \alpha)) \to \alpha\big)$.
• $\vdash \big(\alpha \to (\neg(\neg \alpha))\big)$.

One last thing that I can use is transitivity, which I proved by using the deduction theorem.

• $\{\alpha \to \beta, \, \beta \to \gamma\} \vdash (\alpha \to \gamma)$.

Question. How can I show that $\vdash ((\alpha \to (\neg \alpha)) \to (\neg \alpha))$? It would be nice to let me know your line of thought for tackling this problem.

Answer 1

IMO the "trick" in similar cases of long derivations in Hilbert-style, is to use some intermediate result as Lemma.

Having Ax.3 and the Doble Negation laws, you can prove all Contraposition laws (various combinations).

In addition, it is useful to have an intermediate result: call it

Lemma 1: $(A \to \lnot A) \to (B \to \lnot A)$.

Poof:

1. $(A \to (\lnot A \to \lnot B))$ --- easily proved using Ax.1, Contraposition and Transitivity

2. $(A \to (\lnot A \to \lnot B)) \to ((A \to \lnot A) \to (A \to \lnot B))$ --- Ax.2

3. $((A \to \lnot A) \to (A \to \lnot B))$ --- from 1) and 2) by MP

4. $(A \to \lnot B) \to (B \to \lnot A)$ --- Contraposition

5. $(A \to \lnot A) \to (B \to \lnot A)$ --- from 3) and 4) by Transitivity.

Now for the main result:

1. $(A \to \lnot A) \to ((A \to \lnot A) \to \lnot A)$ --- Lemma 1

2. $[(A \to \lnot A) \to ((A \to \lnot A) \to \lnot A)] \to [((A \to \lnot A) \to (A \to \lnot A)) \to ((A \to \lnot A) \to \lnot A)]$ --- Ax.2

3. $[((A \to \lnot A) \to (A \to \lnot A)) \to ((A \to \lnot A) \to \lnot A) ]$ --- from 1) and 2) by MP

4. $((A \to \lnot A) \to (A \to \lnot A))$ --- from $(\alpha \to \alpha)$

5. $(A \to \lnot A) \to \lnot A$ --- from 3) and 4) by MP.

Answer 2

Expanding on my hint in the comments: define $\top = q \to r \to q$ (where $q, r$ are fresh variables), so that $\top$ is provable by axiom 1. Define $\bot = \neg \top$. Prove $\neg\bot$ using $\top$ and $a \to\neg\neg a$. Prove $\bot \to x$ using axioms 3, 1, MP and $\neg\bot$. Prove $\neg a \to a \to \bot$ using axioms 3 and 1. Prove $(a \to b) \to \neg b \to \neg a$ using axiom 3 and $a \leftrightarrow \neg\neg a$. Prove $(a \to \bot) \to \neg a$ using the previous theorem and $\neg\bot$. Prove $(a \to a \to \bot) \to a \to \bot$ using axioms 1 and 2. Prove the claim using the previous results.

There is most likely a more direct way but I can't be bothered to find it. If you want an elegant proof, stop using Hilbert-style systems.

Answer 3

As you requested...

Screenshot:

Note: Consistent with other infix binary operators in the system used here, '=>' is left-associative.

Text version:

1   A => ~A
    Premise
2   A
    Premise

3   ~A
    Detach, 1, 2

4   A & ~A
    Join, 2, 3


5   ~A
    Conclusion, 2
6   A => ~A => ~A
    Conclusion, 1