On the Propagation of Consistency in Some Systems of Paraconsistent Logic

Abstract

The present paper offers some results on the propagation of consistency in two systems of Logics of Formal Inconsistency(LFIs). One is the system Bk of Avron, which is an extension of the base system mbC of Carnielli, Coniglio and Marcos, and the other is an extension of Bk to the predicate calculus which will be referred to as Bk ^∗. We first present a new characterization of consistency operator in Bk. This reflects the intuition of the consistency operator quite well. Second, we prove that two kinds of propagation of consistency known in the literature are actually equivalent to certain forms of de Morgan’s laws without any occurrence of the consistency operator. Finally, we extend the result in Bk to Bk ^∗.

Appendix

In this appendix, we will provide the readers with full proofs of the theorems which we have omitted in the main body. We begin with the preliminaries for the proofs.

8.1.1 Preliminaries

First, note that several theorems, listed below, can be proved in Bk since it contains axioms from (A1) to (A9) together with the only rule of inference (MP):

$$\displaystyle{ (A \supset (B \supset C)) \supset (B \supset (A \supset C)) }$$

(8.44)

$$\displaystyle{ ((A \wedge B) \supset C) \supset (A \supset (B \supset C)) }$$

(8.45)

$$\displaystyle{ (A \supset (B \supset C)) \supset ((A \wedge B) \supset C) }$$

(8.46)

$$\displaystyle{ (C \supset A) \supset ((C \supset B) \supset (C \supset (A \wedge B))) }$$

(8.47)

$$\displaystyle{ (A \vee (B \wedge C))\!\,\supset \,\!(A \vee B) }$$

(8.48)

$$\displaystyle{ (A \vee (B \wedge C))\!\,\supset \,\!(A \vee C) }$$

(8.49)

$$\displaystyle{ (A\!\,\supset \,\!(B\!\,\supset \,\!(C\!\,\supset \,\!D)))\!\,\supset \,\!((C \wedge A)\!\,\supset \,\!(B\!\,\supset \,\!D) }$$

(8.50)

$$\displaystyle{ (A\!\,\supset \,\!(B\!\,\supset \,\!(C\!\,\supset \,\!D)))\!\,\supset \,\!((B \wedge A)\!\,\supset \,\!(C\!\,\supset \,\!D) }$$

(8.51)

$$\displaystyle{ (A\!\,\supset \,\!(A \wedge B)) \equiv (A\!\,\supset \,\!B) }$$

(8.52)

$$\displaystyle{ (A\!\,\supset \,\!B)\!\,\supset \,\!((B\!\,\supset \,\!A)\!\,\supset \,\!(A \equiv B)) }$$

(8.53)

$$\displaystyle{ ((A\!\,\supset \,\!B)\!\,\supset \,\!B) \equiv (A \vee B) }$$

(8.54)

$$\displaystyle{ ((A\!\,\supset \,\!B)\!\,\supset \,\!A) \equiv A }$$

(8.55)

$$\displaystyle{ (A \vee (B\!\,\supset \,\!C)) \equiv (B\!\,\supset \,\!(C \vee A)) }$$

(8.56)

$$\displaystyle{ (A \wedge B \wedge (C\!\,\supset \,\!A)) \equiv (A \wedge B) }$$

(8.57)

$$\displaystyle{ (A \wedge B \wedge (A \vee C) \wedge D) \equiv (A \wedge D \wedge B) }$$

(8.58)

$$\displaystyle{ B\!\,\supset \,\!(A \equiv (A \wedge B)) }$$

(8.59)

$$\displaystyle{ ((A \vee B)\!\,\supset \,\!C) \equiv ((A\!\,\supset \,\!C) \wedge (B\!\,\supset \,\!C)) }$$

(8.60)

$$\displaystyle{ ((C\!\,\supset \,\!A) \wedge (C\!\,\supset \,\!B)) \equiv (C\!\,\supset \,\!(A \wedge B)) }$$

(8.61)

$$\displaystyle{ (A \equiv B)\!\,\supset \,\!((C\!\,\supset \,\!A) \equiv (C\!\,\supset \,\!B)) }$$

(8.62)

$$\displaystyle{ (A \equiv B)\!\,\supset \,\!((C \wedge A) \equiv (C \wedge B)) }$$

(8.63)

$$\displaystyle{ (A \equiv B)\!\,\supset \,\!((A \vee C) \equiv (B \vee C)) }$$

(8.64)

$$\displaystyle{ (A \equiv B)\!\,\supset \,\!((C \equiv D)\!\,\supset \,\!((A \vee C \vee E) \equiv (B \vee D \vee E)) }$$

(8.65)

$$\displaystyle{ (A \equiv B)\!\,\supset \,\!((C \equiv D)\!\,\supset \,\!((A\!\,\supset \,\!C) \equiv (B\!\,\supset \,\!D))) }$$

(8.66)

$$\displaystyle{ (A \equiv B)\!\,\supset \,\!((C \equiv D)\!\,\supset \,\!((A \wedge C) \equiv (B \wedge D))) }$$

(8.67)

$$\displaystyle{ (A\!\,\supset \,\!(B \equiv C))\!\,\supset \,\!((D\!\,\supset \,\!(E \equiv F))\!\,\supset \,\!((A \wedge D)\!\,\supset \,\!((B \vee E) \equiv (C \vee F)))) }$$

(8.68)

$$\displaystyle{ ((A \wedge B)\!\,\supset \,\!C) \equiv (B\!\,\supset \,\!(A\!\,\supset \,\!C)) }$$

(8.69)

$$\displaystyle{ (((A \vee B) \wedge C)\!\,\supset \,\!D) \equiv (C\!\,\supset \,\!((A\!\,\supset \,\!D) \wedge (B\!\,\supset \,\!D))) }$$

(8.70)

$$\displaystyle{ ((A \wedge B)\!\,\supset \,\!(C \wedge D)) \equiv (B\!\,\supset \,\!((A\!\,\supset \,\!C) \wedge (A\!\,\supset \,\!D))) }$$

(8.71)

$$\displaystyle{ ((A\wedge B) \vee (C \wedge D) \vee (E\!\,\supset \,\!(B \wedge D))) \equiv (E\!\,\supset \,\!((B \vee (C \wedge D)) \wedge (D \vee (A\wedge B)))) }$$

(8.72)

Also note that the following rules of inference can be derived in Bk:

$$\displaystyle{ \frac{\ A\!\,\supset \,\!B\ \ \ \ B\!\,\supset \,\!C\ } {A\!\,\supset \,\!C} }$$

(Syll.)

$$\displaystyle{ \frac{\ A \equiv B\ \ \ \ B \equiv C\ } {A \equiv C} }$$

(EqSyll.)

$$\displaystyle{ \frac{\ A \equiv (B\!\,\supset \,\!C)\ \ \ \ \ C \equiv D\ } {A \equiv (B\!\,\supset \,\!D)} }$$

(8.73)

$$\displaystyle{ \frac{\ A\!\,\supset \,\!(B \equiv (C\!\,\supset \,\!D))\ \ \ \ \ A\!\,\supset \,\!(D \equiv E)\ } {A\!\,\supset \,\!(B \equiv (C\!\,\supset \,\!E))} }$$

(8.74)

$$\displaystyle{ \frac{\ A \equiv (B \wedge C)\ \ \ \ \ B \equiv (D\!\,\supset \,\!E)\ \ \ \ \ C \equiv (D\!\,\supset \,\!F)\ } {A \equiv (D\!\,\supset \,\!(E \wedge F))} }$$

(8.75)

Second, the following theorems which contain negation will be useful.

$$\displaystyle{ (A\!\,\supset \,\!(A \wedge \neg A)) \equiv \neg A }$$

(8.76)

$$\displaystyle{ ((A\!\,\supset \,\!B)\!\,\supset \,\!\neg A) \equiv (A\!\,\supset \,\!(B\!\,\supset \,\!\neg A)) }$$

(8.77)

$$\displaystyle{ ((A\!\,\supset \,\!B)\!\,\supset \,\!\neg B) \equiv \neg B }$$

(8.78)

Finally, as for theorems in the predicate calculus Bk ^∗, note that the following formulas are provable.

$$\displaystyle{ \neg ^{{\ast}}\forall xA(x) \supset \exists x\neg ^{{\ast}}A(x) }$$

(8.79)

$$\displaystyle{ \neg ^{{\ast}}\forall xA(x) \equiv \exists x\neg ^{{\ast}}A(x) }$$

(8.80)

$$\displaystyle{ \neg ^{{\ast}}\exists xA(x) \supset \forall x\neg ^{{\ast}}A(x) }$$

(8.81)

$$\displaystyle{ \neg ^{{\ast}}\exists xA(x) \equiv \forall x\neg ^{{\ast}}A(x) }$$

(8.82)

$$\displaystyle{ (\forall xA(x) \vee B) \equiv \forall x(A(x) \vee B) }$$

(8.83)

Here, of course, x is not free in B of (8.83). Also, note that the following rules of inference can be derived in Bk ^∗:

$$\displaystyle{ \frac{A(x) \supset B(x)} {\forall xA(x) \supset \forall xB(x)} }$$

(8.84)

$$\displaystyle{ \frac{A(x) \supset B(x)} {\exists xA(x) \supset \exists xB(x)} }$$

(8.85)

$$\displaystyle{ \frac{A(x) \equiv B(x)} {\exists xA(x) \equiv \exists xB(x)} }$$

(8.86)

$$\displaystyle{ \frac{A(x) \equiv (B(x) \wedge C(x))} {\forall xA(x) \equiv (\forall xB(x) \wedge \forall xC(x))} }$$

(8.87)

$$\displaystyle{ \frac{A\!\,\supset \,\!(B(x) \equiv C(x))} {A\!\,\supset \,\!(\exists xB(x) \equiv \exists xC(x))} }$$

(8.88)

Here, of course, x is not free in A of (8.88). We shall make use of the above theorems and rules of inference in the following proofs.

Proof of Theorem 1.

For (8.1) and (8.2), just apply (8.48) and (8.49) to (k). For (8.3), we obtain $(A \vee \neg A)\!\,\supset \,\!(\neg ^{{\ast}}A\!\,\supset \,\!\neg A)$ by (CN) and therefore $\neg ^{{\ast}}A\!\,\supset \,\!\neg A$ follows by making use of (A11) and (MP). For (8.4):

Table 1

For (8.5), it immediately follows from (8.4) by (CN) and as for (8.6), just note that we have (8.4) together with $\neg ^{{\ast}}(A \wedge \neg A) \equiv (\neg A\!\,\supset \,\!\neg ^{{\ast}}A)$ which can be obtained by (CN). For (8.7), note that we obtain $(\neg A\!\,\supset \,\!\neg ^{{\ast}}A) \equiv (\neg A \equiv \neg ^{{\ast}}A)$ by (8.3) and (8.59), and combine it with (8.6). For the proof of (8.8), note that we obtain $(\neg A \equiv \neg ^{{\ast}}A) \equiv (\neg ^{{\ast}}\neg A \equiv A)$ by (CN). For (8.9):

Table 2

For (8.10):

Table 3

This completes the proof.

Proof of Proposition 5.

Proof proceeds analogously for all four formulas. We shall therefore only prove two cases, for (8.11) and (8.12).

For (8.11):

Table 4

For (8.12):

Table 5

This completes the proof.

Proof of Theorem 5.

For (8.19):

Table 6

For (8.20): We abbreviate the formula $\neg (A \wedge B)\!\,\supset \,\!(\neg A \vee \neg B)$ to X.

Table 7

For (8.21):

Table 8

For (8.22): Proof is analogous to the proof for (8.21); indeed, we just need to replace $\neg (A \vee B),(\neg A \wedge \neg B)$ with $\neg (A\!\,\supset \,\!B),(A \wedge \neg B)$ respectively.

This completes the proof.

Proof of Proposition 6.

For (8.27):

Table 9

For (8.28):

Table 10

For (8.29):

Table 11

For (8.30):

Table 12

This completes the proof.

Proof of Theorem 7.

For (8.31):

Table 13

For (8.32):

Table 14

For (8.33):

Table 15

This completes the proof.

Proof of Lemma 5.

Since the proofs for two formulas (8.34) and (8.35) are analogous, we only prove the former whose proof runs as follows:

Table 16

This completes the proof.

Proof of Theorem 9.

For (8.38):

Table 17

For (8.39):

Table 18

This completes the proof.

Proof of Theorem 11.

For (8.42):

Table 19

For (8.43):

Table 20

This completes the proof.