# Counterexample equivalent in relevant logic DL

On page 7 in the article referred to below an axiom $$D9$$ is stated as follows: $$A\to B\to.\lnot(A \& \lnot B)~\\ (\text{equivalently: } (A\to\lnot A)\to\lnot A)$$

How may one prove the alleged equivalence, using the rest of the axioms and the inference rules:

D1: $$\vdash A\to A$$

D2: $$\vdash((A\to B)\wedge(B\to C))\to(A\to C)$$

D3: $$\vdash(A\wedge B)\to A$$

D4: $$\vdash(A\wedge B)\to B$$

D5: $$\vdash((A\to B)\wedge (A\to C))\to(A\to (B\wedge C))$$

D6: $$\vdash(A\wedge(B\vee C))\to ((A\wedge B)\vee(A\vee C))$$

D7: $$\vdash\lnot\lnot A\to A$$

D8: $$\vdash(A\to\lnot B)\to(B\to\lnot A)$$

R1: $$\vdash A \ \& \vdash (A\to B) \Rightarrow \ \vdash B$$

R2: $$\vdash A \ \& \vdash B \Rightarrow \ \vdash A\wedge B$$

R3: $$\vdash A\to B \ \& \ \vdash C\to D\ \Rightarrow \ \vdash(B\to C)\to (A\to D)$$

Dialectical Logic, Classical Logic, and the Consistency of the World Author(s): Richard Routley and Robert K. Meyer Source: Studies in Soviet Thought , Jun., 1976, Vol. 16, No. 1/2 (Jun., 1976), pp. 1-25. Link (behind JSTOR paywall)

• What are the other axioms of the relevant system? (Especially given that this is a less-well-known system and the paper's behind a paywall.) – Noah Schweber Oct 28 '20 at 19:33
• @NoahSchweber I now added the remaining axioms and inference rules from the publication. – Frode Alfson Bjørdal Oct 28 '20 at 20:27

## 1 Answer

From D3 and D4,

1- $$A\wedge\neg B\rightarrow A$$.

2- $$A\wedge\neg B\rightarrow \neg B$$.

From 2, D8 and R1,

3- $$B\rightarrow \neg(A\wedge\neg B)$$.

From 1, 3 and R3 (taking $$B=A$$, $$C=B$$, $$A=A\wedge\neg B$$ and $$D=\neg(A\wedge\neg B)$$)

4- $$(A\rightarrow B)\rightarrow (A\wedge\neg B\rightarrow \neg(A\wedge\neg B))$$.

From the axiom $$(A\rightarrow \neg A)\rightarrow A)$$,

5- $$(A\wedge\neg B\rightarrow\neg(A\wedge\neg B))\rightarrow\neg(A\wedge\neg B)$$.

From 4, 5, by transitivity of implication (D2, R1, R2),

6- $$(A\rightarrow B)\rightarrow \neg(A\wedge\neg B)$$.

The converse implication is simpler. Take $$B=\neg A$$ in $$(A\rightarrow B)\rightarrow \neg(A\wedge\neg B)$$. Then,

1- $$(A\rightarrow \neg A)\rightarrow \neg(A\wedge\neg\neg A)$$.

From D5, using the double negation law (easily obtained from D8 and R1)

2- $$A\rightarrow (A\wedge\neg\neg A)$$, so $$A\rightarrow\neg\neg(A\wedge\neg\neg A)$$ (the double negation law once more.)

From D8,

3- $$(A \rightarrow\neg\neg(A\wedge\neg\neg A))\rightarrow (\neg(A\wedge\neg\neg A)\rightarrow\neg A))$$.

From 2, 3 and R1,

4- $$\neg(A\wedge\neg\neg A)\rightarrow \neg A$$.

The conclusion follows from 1, 4 and the transitivity of implication.