How to prove $A \lor \neg A$ in an axiom system for Relevance Logic R

Question

Relevance Logic $\sf R$ is axiomatized over the following axioms and rules:

1. $A \to A$
2. $(A \to B) \to ((B \to C) \to (A \to C))$
3. $A \to ((A \to B) \to B)$
4. $(A \to (A \to B)) \to (A \to B)$
5. $(A \land B) \to A$
6. $(A \land B) \to B$
7. $((A \to B) \land (A \to C)) \to (A \to (B \land C))$
8. $((A \lor B) \to C) \leftrightarrow ((A \to C) \land (B \to C))$
9. $(A \to \neg B) \to (B \to \neg A)$
10. $\neg \neg A \to A$

MP: $A, (A \to B) \vdash B$

Adj: $A,B \vdash (A \land B)$

Also, relevance logic defines $(A \lor B)$ as $\neg (\neg A \land \neg B)$, and $(A \leftrightarrow B)$ as $((A \to B) \land (B \to A))$.

I’ve been puzzling for a few hours about how to prove $(A \lor \neg A)$ in $\sf R$. I realize that if I could prove any of the following, then I could complete the proof:

$(A \to \neg A) \to \neg A$

$(A \to B) \to ((A \to \neg B) \to \neg A)$

$(\neg A \to A) \to A$

etc.

I do realize that if I could prove $(A \to B) \to ((A \to \neg B) \to \neg A)$, then I could prove the others, including LEM

I also realize that per the Wikipedia article on Relevance Logic, there are weaker systems that validate $(A \lor \neg A)$; specifically, I don’t think axiom three is necessary for the proof, since it’s not present in relevance logic $\sf T$, which still proves LEM.

If anyone could give me a hint as to how to proceed in proving any of the above, I would be quite grateful.

Answer 1

Though Graham Priest and Edwin Mares are my main sources, a tidy collection of the definitions employed in the following presentation can be found in Shawn Standefer's open access paper "Routes to Relevance: Philosophies of Relevant Logics."

Starting with the weakest system B ("Base Logic"), we build up systems as depicted in the diagram below (from Stephen Read's Relevant Logic: A Philosophical Examination of Inference):

So, the Axiom Schemas:

1. $A\rightarrow A\tag{Identity}$

2. $A\rightarrow(A\vee B);\;B\rightarrow(A\vee B)\tag{$\vee$-Introduction}$

3. $((A\rightarrow B)\wedge (A\rightarrow C))\rightarrow(A\rightarrow(B\wedge C))\tag{$\wedge$-Introduction}$

4. $((A\rightarrow C)\wedge(B\rightarrow C))\rightarrow((A\vee B) \rightarrow C)\tag{$\vee$-Elimination}$

5. $(A\wedge B)\rightarrow A;\;(A\wedge B)\rightarrow B\tag{$\wedge$-Elimination}$

6. $(A\wedge(B\vee C))\rightarrow((A\wedge B)\vee(A\wedge C))\tag{Distribution}$

7. $\neg\neg A\rightarrow A\tag{Double Negation-Elimination}$

8. $(A\rightarrow\neg B)\rightarrow(B\rightarrow\neg A)\tag{Contrapositive}$

and the Inference Rules:

1. $A\rightarrow B, A\vdash B\tag{Modus Ponens}$

2. $A\rightarrow B\vdash(B\rightarrow C)\rightarrow(A\rightarrow C)\tag{Suffixing}$

3. $B\rightarrow C\vdash(A\rightarrow B)\rightarrow(A\rightarrow C)\tag{Prefixing}$

4. $A\rightarrow\neg B\vdash B\rightarrow\neg A\tag{Contraposition}$

5. $(A\rightarrow(B\rightarrow C))\rightarrow(B\rightarrow(A\rightarrow C))\tag{Permutation}$

For the system R of relevant implication, we add:

$$(A\rightarrow(A\rightarrow B))\rightarrow(A\rightarrow B)\tag{Contraction}$$

Note that double-negation introduction (DN-Introduction) can be derived as as a theorem:

1. $\neg A\rightarrow\neg A\tag{Identity}$
2. $A\rightarrow\neg\neg A\tag{Contraposition 1}$

As a lemma, so to say, we need to show that (an alternative form of) consequentia mirabilis is derivable:

1. $(A\rightarrow\neg A)\rightarrow(A\rightarrow\neg A)\tag{Identity}$
2. $A\rightarrow((A\rightarrow\neg A)\rightarrow\neg A)\tag{Permutation 1}$
3. $((A\rightarrow\neg A)\rightarrow\neg A)\rightarrow(A\rightarrow\neg(A \rightarrow \neg A))\tag{Contrapositive}$
4. $A\rightarrow((A\rightarrow\neg A)\rightarrow\neg A)\rightarrow (A\rightarrow(A\rightarrow\neg(A\rightarrow\neg A)))\tag{Prefixing}$
5. $A\rightarrow(A\rightarrow\neg(A\rightarrow\neg A))\tag{Modus ponens 2, 4}$
6. $(A\rightarrow (A\rightarrow\neg(A\rightarrow\neg A)))\rightarrow(A\rightarrow\neg(A\rightarrow\neg A))\tag{Contraction}$
7. $A\rightarrow\neg(A\rightarrow\neg A)\tag{Modus ponens 5, 6}$
8. $(A\rightarrow\neg A)\rightarrow\neg A\tag{Contraposition 7}$

Then, we can work out the proof:

1. $A \rightarrow (A \vee \neg A)\tag{$\vee$-Introduction}$
2. $(A\vee\neg A)\rightarrow\neg\neg(A\vee\neg A)\tag{DN-Introduction}$
3. $A\rightarrow\neg\neg(A\vee\neg A)\tag{Hypothetical syllogism 1, 2}$
4. $\neg(A\vee\neg A)\rightarrow\neg A\tag{Contraposition 3}$
5. $\neg A \rightarrow (A \vee \neg A)\tag{$\vee$-Introduction}$
6. $\neg(A\vee\neg A)\rightarrow(A\vee\neg A)\tag{Hypothetical syllogism 4, 5}$
7. $\neg(A\vee\neg A)\rightarrow\neg\neg(A\vee\neg A)\tag{Hypothetical syllogism 2, 6}$
8. $(\neg (A\vee\neg A)\rightarrow\neg\neg(A\vee\neg A))\rightarrow\neg\neg (A \vee\neg A)\tag{Consequentia mirabilis}$
9. $\neg\neg(A\vee\neg A)\tag{Modens ponens 7, 8}$
10. $\neg\neg(A\vee\neg A)\rightarrow(A\vee\neg A)\tag{DN-Elimination}$
11. $A\vee\neg A\tag{Modens ponens 9, 10}$