What is the proof of reductio ad absurdum (RAA) in a Łukasiewicz axiom system for propositional logic with only modus ponens?

Question

I am working with the following Lukasiewicz axiom system:

Axiom Schema 1: $\alpha \rightarrow (\beta \rightarrow \alpha )$

Axiom Schema 2: $(\alpha \rightarrow (\beta \rightarrow \gamma)) \rightarrow ((\alpha \rightarrow \beta) \rightarrow (\alpha \rightarrow \gamma)) $

Axiom Schema 3: $(\lnot \alpha \rightarrow \lnot \beta) \rightarrow (\beta \rightarrow \alpha)$

Rule of Inference (Modus ponens): From $\alpha$ and $\alpha \rightarrow \beta$, infer $\beta$.

I am looking for a proof of the following:

Reductio ad absurdum (RAA): If $\Gamma \cup \{ \alpha \}$ is inconsistent, then $\Gamma ⊢ \lnot \alpha$.

I have seen a proof of the first-order case, but unless I am mistaken it relies on the deduction theorem for first-order logic and the completeness theorem for propositional logic, so I am trying to find a proof of the propositional case which does not depend on the completeness theorem for propositional logic. The deduction theorem can (and I am assuming must?) be assumed.

Proofs with similar axiom systems (using only $\lnot$ and $\rightarrow$) are fine, but slightly less ideal.

References, suggestions, and edits are welcomed.

Answer 1

So you want to prove the following theorem:

Theorem: If $\Gamma,\phi \vdash \psi$ and $\Gamma, \phi \vdash \neg \psi$, then $\Gamma \vdash \neg \phi$

Proof:

First, I'll assume that you can use the Deduction Theorem, which states that for any $\Gamma$, $\varphi$, and $\psi$:

If $\Gamma \cup \{ \varphi \} \vdash \psi$, then $\Gamma \vdash \varphi \rightarrow \psi$

So if $\Gamma,\phi \vdash \psi$ and $\Gamma, \phi \vdash \neg \psi$, then by the Deduction Theorem we have $\Gamma \vdash \phi \to \psi$ and $\Gamma \vdash \phi \to \neg \psi$

This means that if can show that $\phi \to \psi, \phi \to \neg \psi \vdash \neg \phi$, then we're there.

This is not easy, but here goes:

First, let's prove: $\phi \to \psi, \psi \to \chi, \phi \vdash \chi$:

\begin{array}{lll} 1&\phi \to \psi & Premise\\ 2& \psi \to \chi & Premise\\ 3&\phi& Premise\\ 4&\psi& MP \ 1,3\\ 5&\chi& MP \ 2,4\\ \end{array}

By the Deduction Theorem, this gives us Hypothetical Syllogism (HS): $\phi \to \psi, \psi \to \chi \vdash \phi \to \chi$

Now let's prove the general principle that $\neg \phi \vdash (\phi \to \psi)$:

\begin{array}{lll} 1. &\neg \phi& Premise\\ 2. &\neg \phi \to (\neg \psi \to \neg \phi)& Axiom \ 1\\ 3. &\neg \psi \to \neg \phi& MP \ 1,2\\ 4. &(\neg \psi \to \neg \phi) \to (\phi \to \psi)& Axiom \ 3\\ 5. &\phi \to \psi& MP \ 3,4\\ \end{array}

With the Deduction Theorem, this means $\vdash \neg \phi \to (\phi \to \psi)$ (Duns Scotus Law)

Let's use Duns Scotus to show that $\neg \phi \to \phi \vdash \phi$

\begin{array}{lll} 1. &\neg \phi \to \phi& Premise\\ 2. &\neg \phi \to (\phi \to \neg (\neg \phi \to \phi))& Duns \ Scotus\\ 3. &(\neg \phi \to (\phi \to \neg (\neg \phi \to \phi))) \to ((\neg \phi \to \phi) \to (\neg \phi \to \neg (\neg \phi \to \phi)))& Axiom \ 2\\ 4. &(\neg \phi \to \phi) \to (\neg \phi \to \neg (\neg \phi \to \phi))& MP \ 2,3\\ 5. &\neg \phi \to \neg (\neg \phi \to \phi)& MP \ 1,4\\ 6. &(\neg \phi \to \neg (\neg \phi \to \phi)) \to ((\neg \phi \to \phi) \to \phi)& Axiom \ 3\\ 7. &(\neg \phi \to \phi) \to \phi& MP \ 5,6\\ 8. &\phi& MP \ 1,7\\ \end{array}

By the Deduction Theorem, this means $\vdash (\neg \phi \to \phi) \to \phi$ (Law of Clavius)

Using Duns Scotus and the Law of Clavius, we can now show that $ \neg \neg \phi \vdash \phi$:

\begin{array}{lll} 1. &\neg \neg \phi& Premise\\ 2. &\neg \neg \phi \to (\neg \phi \to \phi)& Duns \ Scotus\\ 3. &\neg \phi \to \phi& MP \ 1,2\\ 4. &(\neg \phi \to \phi) \to \phi& Clavius\\ 5. &\phi& MP \ 3,4\\ \end{array}

By the Deduction Theorem, this also means that $\vdash \neg \neg \phi \to \phi$ (DN Elim or DNE)

Finally, we can show the desired $\phi \to \psi, \phi \to \neg \psi \vdash \neg \phi$:

\begin{array}{lll} 1. &\phi \to \psi& Premise\\ 2. &\phi \to \neg \psi& Premise\\ 3. &\neg \neg \phi \to \phi& DNE\\ 4. &\neg \neg \phi \to \psi& HS \ 1,3\\ 5. &\neg \neg \phi \to \neg \psi& HS \ 2,3\\ 6. &(\neg \neg \phi \to \neg \psi) \to (\psi \to \neg \phi)& Axiom \ 3\\ 7. &\psi \to \neg \phi& MP \ 5,6\\ 8. &\neg \neg \phi \to \neg \phi& HS \ 4,7\\ 9. &(\neg \neg \phi \to \neg \phi) \to \neg \phi& Clavius\\ 10. &\neg \phi& MP \ 8,9\\ \end{array}

Now, you can actually get a little more quickly to $\neg \neg \phi \vdash \phi$ as follows:

\begin{array}{lll} 1&\neg \neg \phi&Premise\\ 2&\neg \neg \phi \to (\neg \neg \neg \neg \phi \to \neg \neg \phi)&Axiom \ 1\\ 3&\neg \neg \neg \neg \phi \to \neg \neg \phi&MP \ 1,2\\ 4&(\neg \neg \neg \neg \phi \to \neg \neg \phi) \to (\neg \phi \to \neg \neg \neg \phi) & Axiom \ 3\\ 5& \neg \phi \to \neg \neg \neg \phi & MP \ 3,4\\ 6&(\neg \phi \to \neg \neg \neg \phi) \to (\neg \neg \phi \to \phi) & Axiom \ 3\\ 7& \neg \neg \phi \to \phi & MP \ 5,6\\ 8&\phi&MP \ 1,7\\ \end{array}

However, since the proof of $\phi \to \psi, \phi \to \neg \psi \vdash \neg \phi$ relies on Clavius, I took the road that I did.