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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)

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    $\begingroup$ 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.) $\endgroup$ – Noah Schweber Oct 28 '20 at 19:33
  • $\begingroup$ @NoahSchweber I now added the remaining axioms and inference rules from the publication. $\endgroup$ – Frode Alfson Bjørdal Oct 28 '20 at 20:27
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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.

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