A Paraconsistent Logic Obtained from an Algebra-Valued Model of Set Theory

Abstract

This paper presents a three-valued paraconsistent logic obtained from some algebra-valued model of set theory. Soundness and completeness theorems are established. The logic has been compared with other three-valued paraconsistent logics.

Keywords

Mathematics Subject Classification (2000)

We are thankful to Prof. Benedikt Löwe for his support in the set theoretic investigations indicated in this paper. We also acknowledge the anonymous referee for making valuable comments. The first author’s research is funded by CSIR Ph.D. fellowship, Govt. of India, grant number: 09/028(0754)/2009 -EMR-1.

Appendices

Appendix A

Proof of the Lemma 7.3.5:

As it was indicated that the proof will be by induction on the complexity of $\varphi$. Let, $\Gamma = \{p_{i_1}', p_{i_2}', \ldots , p_{i_k}'\}$.

Basis step: It is obvious when the complexity is 0.

Induction hypothesis: Assume the lemma holds well for formulas with complexity less than n.

Induction step: Let the complexity of $\varphi$ be n.

Case 1: Let $\varphi = \lnot \psi$.

Clearly, the complexity of $\psi$ is less than n and the propositional letters in $\psi$ are exactly same as the propositional letters in $\varphi$.

Subcase 1.1: If $v(\psi ) = 1$ then $v(\varphi ) = 0$. Hence, by our construction, $\psi ' = \psi \wedge (\lnot \psi \rightarrow \bot )$ and $\varphi ' = \lnot \varphi \wedge (\varphi \rightarrow \bot )$. Here we get

$\Gamma \; \vdash$

1. $\psi '$       induction hypothesis

            2. $\psi$       Ax3 and M.P.

            3. $\lnot \psi \rightarrow \bot$       Ax4 and M.P.

            4. $\psi \rightarrow \lnot \lnot \psi$       Ax8

            5. $\lnot \lnot \psi$       M.P. 2, 4

            6. $\lnot \varphi \wedge (\varphi \rightarrow \bot )$       Rule 1 on 3 and 5

Hence, in Subcase 1.1, $\Gamma \; \vdash \; \varphi '$.

Subcase 1.2: If then . So by the construction, $\psi ' = \psi \wedge \lnot \psi$ and $\varphi ' = \varphi \wedge \lnot \varphi$. So we have

       $\Gamma \; \vdash$

            1. $\psi '$       induction hypothesis

            2. $\psi$       Ax3 and M.P.

            3. $\lnot \psi$       Ax4 and M.P.

            4. $\psi \rightarrow \lnot \lnot \psi$       Ax8

            5. $\lnot \lnot \psi$       M.P. 2, 4

            6. $\varphi \wedge \lnot \varphi$       Rule 1 on 3 and 5

Hence, $\Gamma \; \vdash \; \varphi '$ holds here.

Subcase 1.3: If $v(\psi ) = 0$ then $v(\varphi ) = 1$. Hence,

                        $\psi ' = \lnot \psi \wedge (\psi \rightarrow \bot )$ and $\varphi ' = \varphi \wedge (\lnot \varphi \rightarrow \bot )$.

The following derivation can be made

       $\Gamma \; \vdash$

            1. $\psi '$       induction hypothesis

            2. $\lnot \psi$       Ax3 and M.P.

            3. $\psi \rightarrow \bot$       Ax4 and M.P.

            4. $(\lnot \lnot \psi \rightarrow \psi ) \rightarrow [(\psi \rightarrow \bot ) \rightarrow (\lnot \lnot \psi \rightarrow \bot )]$       Theorem 7.3.4(ii)

            5. $\lnot \lnot \psi \rightarrow \psi$       Ax8

            6. $(\psi \rightarrow \bot ) \rightarrow (\lnot \lnot \psi \rightarrow \bot )$       M.P. 4, 5

            7. $\lnot \lnot \psi \rightarrow \bot$       M.P. 3, 6

            8. $\varphi '$       Rule 1 on 2, 7

Hence, in Case 1 we always get $\Gamma \; \vdash \; \varphi '$.

Case 2: Let $\varphi = \psi \wedge \gamma$.

Obviously, both the complexities of $\psi$ and $\gamma$ are less than n and the sets of propositional letters in $\varphi$ and $\psi$ are proper subsets of $\{p_{i_1}', p_{i_2}', \ldots , p_{i_k}'\}$, the set of propositional letters in $\varphi$. Hence, clearly by the induction hypothesis and monotonicity property we get

$$\begin{aligned} \Gamma \; \vdash \; \psi ' \quad \text {and} \quad \Gamma \; \vdash \; \gamma '. \end{aligned}$$

Subcase 2.1: If any one of $v(\psi )$ and $v(\gamma )$ is 0 then it can be proved $\Gamma \; \vdash \; \varphi '$. Without loss of generality, let $v(\psi ) = 0$, then $v(\varphi ) = 0$. Hence, we get the following:

$$\begin{aligned} \psi ' = \lnot \psi \wedge (\psi \rightarrow \bot ) \quad \text {and} \quad \varphi ' = \lnot \varphi \wedge (\varphi \rightarrow \bot ). \end{aligned}$$

Since $\Gamma \; \vdash \; \psi '$,

       $\Gamma \; \vdash$

            1. $\lnot \psi$       Ax3 and M.P.

            2. $\psi \rightarrow \bot$       Ax4 and M.P.

            3. $\lnot \psi \rightarrow (\lnot \psi \vee \lnot \gamma )$       Ax5

            4. $\lnot \psi \vee \lnot \gamma$       M.P. 1, 3

            5. $(\lnot \psi \vee \lnot \gamma ) \rightarrow \lnot (\psi \wedge \gamma )$       Ax9

            6. $\lnot (\psi \wedge \gamma )$    M.P. 4, 5

            7. $(\psi \rightarrow \bot ) \rightarrow (\psi \wedge \gamma \rightarrow \bot )$       Theorem 7.3.4(i)

            8. $\psi \wedge \gamma \rightarrow \bot$       M.P. 2, 7

            9. $\varphi '$       Rule 1 on 6, 8

Subcase 2.2: If and then also. So by the definition,

$$\begin{aligned} \psi ' = \psi \wedge \lnot \psi , \quad \gamma ' = \gamma \wedge \lnot \gamma \quad \text {and} \quad \varphi ' = \varphi \wedge \lnot \varphi . \end{aligned}$$

Now for proving $\Gamma \; \vdash \; \varphi '$, i.e., $\Gamma \; \vdash \; (\psi \wedge \gamma ) \wedge \lnot (\psi \wedge \gamma )$ we go through the following derivation, using $\Gamma \; \vdash \; \psi '$ and $\Gamma \; \vdash \; \gamma '$.

$\Gamma \; \vdash$

1. $\psi$       Ax3 and M.P.

2. $\lnot \psi$       Ax4 and M.P.

3. $\gamma$       Ax3 and M.P.

4. $\psi \wedge \gamma$       Rule 1 on 1, 3

5. $\lnot \psi \rightarrow (\lnot \psi \vee \lnot \gamma )$       Ax5

6. $(\lnot \psi \vee \lnot \gamma )$       M.P. 2, 5

7. $(\lnot \psi \vee \lnot \gamma ) \rightarrow \lnot (\psi \wedge \gamma )$       Ax9

8. $\lnot (\psi \wedge \gamma )$       M.P. 6, 7

9. $\varphi '$       Rule 1 on 4, 8

Subcase 2.3: If and $v(\gamma ) = 1$ then also. Hence,

$$\begin{aligned} \psi ' = \psi \wedge \lnot \psi , \quad \gamma ' = \gamma \wedge (\lnot \gamma \rightarrow \bot ) \quad \text {and} \quad \varphi ' = \varphi \wedge \lnot \varphi . \end{aligned}$$

Since $\Gamma \; \vdash \; \psi '$ and $\Gamma \; \vdash \; \gamma '$, using Axiom 1, 2 and rule M.P. we get

$$\begin{aligned} \Gamma \; \vdash \; \psi , \quad \Gamma \; \vdash \; \lnot \psi \quad \text {and} \quad \Gamma \; \vdash \; \gamma . \end{aligned}$$

Now following the same derivation as above we can prove $\Gamma \; \vdash \; \psi '$.

Subcase 2.4: If $v(\psi ) = 1$ and $v(\gamma ) = 1$ then $v(\varphi ) = 1$. Therefore, by our construction,

$$\begin{aligned} \psi ' = \psi \wedge (\lnot \psi \rightarrow \bot ), \quad \gamma ' = \gamma \wedge (\lnot \gamma \rightarrow \bot ) \quad \text {and} \quad \varphi ' = \varphi \wedge (\lnot \varphi \rightarrow \bot ). \end{aligned}$$

So we have to prove $\Gamma \; \vdash \; \varphi '$, i.e., $\Gamma \; \vdash \; (\psi \wedge \gamma ) \wedge [\lnot (\psi \wedge \gamma ) \rightarrow \bot ]$. The derivation is as follows:

$\Gamma \; \vdash$

1. $\psi$       Ax3 and M.P.

2. $\gamma$       Ax3 and M.P.

3. $\lnot \psi \rightarrow \bot$       Ax4 and M.P.

4. $\lnot \gamma \rightarrow \bot$       Ax4 and M.P.

5. $\psi \wedge \gamma$       Rule 1 on 1 and 2

6. $(\lnot \psi \rightarrow \bot ) \wedge (\lnot \gamma \rightarrow \bot )$       Rule 1 on 3 and 4

7. $(\lnot \psi \rightarrow \bot ) \wedge (\lnot \gamma \rightarrow \bot ) \rightarrow (\lnot \psi \vee \lnot \gamma \rightarrow \bot )$       Ax6

8. $\lnot \psi \vee \lnot \gamma \rightarrow \bot$       M.P. 6, 7

9. $\lnot (\psi \wedge \gamma ) \rightarrow (\lnot \psi \vee \lnot \gamma )$       Ax9

10. $[\lnot (\psi \wedge \gamma ) \rightarrow (\lnot \psi \vee \lnot \gamma )] \rightarrow$

$[((\lnot \psi \vee \lnot \gamma ) \rightarrow \bot ) \rightarrow (\lnot (\psi \wedge \gamma ) \rightarrow \bot )]$       Theorem 7.3.4(ii)

11. $\lnot (\psi \wedge \gamma ) \rightarrow \bot$ M.P. repeatedly on 10, 9, 8

12. $\varphi '$       Rule 1 on 5, 11

Subcase 2.5: If $v(\psi ) = 1$ and then by the same derivation in Subcase 2.3 it can be proved $\Gamma \; \vdash \; \varphi '$.

Hence, in Case 2 we can always prove $\Gamma \; \vdash \; \varphi '$.

Case 3: Let $\varphi = \psi \vee \gamma$.

Since $\varphi \vee \psi$ can be abbreviated as $\lnot (\lnot \varphi \wedge \lnot \psi )$ therefore using Case 1 and Case 2, $\Gamma \; \vdash \; \varphi '$ can be proved in this case also.

Case 4: Let $\varphi = \psi \rightarrow \gamma$.

Obviously, both the complexities of $\psi$ and $\gamma$ are less than n and the sets of propositional letters in $\varphi$ and $\psi$ are subsets of $\{p_{i_1}', p_{i_2}', \ldots , p_{i_k}'\}$, the set of propositional letters in $\varphi$. Hence, clearly by the induction hypothesis and monotonicity property we get

$$\begin{aligned} \Gamma \; \vdash \; \psi ' \quad \text {and} \quad \Gamma \; \vdash \; \gamma '. \end{aligned}$$

Subcase 4.1: If $v(\gamma ) = 1$ then no matter what $v(\psi )$ is, $v(\varphi ) = 1$ always. So we get $\gamma ' = \gamma \wedge (\lnot \gamma \rightarrow \bot )$ and $\varphi ' = \varphi \wedge (\lnot \varphi \rightarrow \bot ) = (\psi \rightarrow \gamma ) \wedge [\lnot (\psi \rightarrow \gamma ) \rightarrow \bot ].$

Now for proving $\Gamma \; \vdash \; \varphi '$ we go through the following derivation:

$\Gamma \; \vdash$

1. $\gamma$       Ax3 and M.P.

2. $\lnot \gamma \rightarrow \bot$       Ax4 and M.P.

3. $\gamma \rightarrow (\psi \rightarrow \gamma )$       Ax1

4. $\psi \rightarrow \gamma$       M.P. 1, 3

5. $(\lnot \gamma \rightarrow \bot ) \rightarrow [\lnot (\psi \rightarrow \gamma ) \rightarrow \bot ]$       Ax12

6. $\lnot (\psi \rightarrow \gamma ) \rightarrow \bot$       M.P. 2, 5

7. $\varphi '$       Rule 1 on 4, 6

Subcase 4.2: If then always $v(\varphi ) = 1$. Hence, by the definition,

$$\begin{aligned} \gamma ' = \gamma \wedge \lnot \gamma \quad \text {and} \quad \varphi ' = \varphi \wedge (\lnot \varphi \rightarrow \bot ) = (\psi \rightarrow \gamma ) \wedge [\lnot (\psi \rightarrow \gamma ) \rightarrow \bot ]. \end{aligned}$$

Hence, we get the following:

$\Gamma \; \vdash$

1. $\gamma \wedge \lnot \gamma$       induction hypothesis

2. $\gamma$       Ax3 and M.P.

3. $\gamma \rightarrow (\psi \rightarrow \gamma )$       Ax1

4. $\psi \rightarrow \gamma$       M.P. 1, 2

5. $(\gamma \wedge \lnot \gamma ) \rightarrow [\lnot (\psi \rightarrow \gamma ) \rightarrow \bot ]$       Ax10

6. $\lnot (\psi \rightarrow \gamma ) \rightarrow \bot$       M.P. 1, 5

7. $\varphi '$       Rule 1 on 3, 6

Subcase 4.3: If and $v(\psi ) = 0$ then $v(\varphi ) = 1$. So by the construction, $\gamma ' = \lnot \gamma \wedge (\gamma \rightarrow \bot )$, $\psi ' = \lnot \psi \wedge (\psi \rightarrow \bot )$ and

$$\begin{aligned} \varphi '&= \varphi \wedge (\lnot \varphi \rightarrow \bot ) \\&= (\psi \rightarrow \gamma ) \wedge [\lnot (\psi \rightarrow \gamma ) \rightarrow \bot ]. \end{aligned}$$

Now the following derivation shows that $\Gamma \; \vdash \; \varphi '$ holds in this subcase also.

$\Gamma \; \vdash$

1. $\psi \rightarrow \bot$       Ax4 and M.P.

2. $\bot \rightarrow \gamma$       Ax13

3. $(\psi \rightarrow \bot ) \rightarrow [(\bot \rightarrow \gamma ) \rightarrow (\psi \rightarrow \gamma )]$       Theorem 7.3.4(ii)

4. $(\bot \rightarrow \gamma ) \rightarrow (\psi \rightarrow \gamma )$       M.P. 1, 3

5. $\psi \rightarrow \gamma$       M.P. 2, 4

6. $(\psi \rightarrow \bot ) \rightarrow [\lnot (\psi \rightarrow \gamma ) \rightarrow \bot ]$       Ax11

7. $\lnot (\psi \rightarrow \gamma ) \rightarrow \bot$       M.P. 1, 6

8. $\varphi '$       Rule 1 on 5, 7

Subcase 4.4: If $v(\gamma ) = 0$ and $v(\psi ) = 1$ then $v(\varphi ) = 0$. Therefore, $\gamma ' = \lnot \gamma \wedge (\gamma \rightarrow \bot )$, $\psi ' = \psi \wedge (\lnot \psi \rightarrow \bot )$ and

$$\begin{aligned} \varphi '&= \lnot \varphi \wedge (\varphi \rightarrow \bot ) \\&= \lnot (\psi \rightarrow \gamma ) \wedge [(\psi \rightarrow \gamma ) \rightarrow \bot ]. \end{aligned}$$

Deduction theorem will be used here for proving $\Gamma \; \vdash \; \varphi '$. Since we know $\Gamma \; \vdash \; \psi '$ and $\Gamma \; \vdash \; \gamma '$

$\Gamma \cup \{\psi \rightarrow \gamma \} \; \vdash$

1. $\psi '$       monotonicity

2. $\psi$       Ax3 and M.P. with 1

3. $\gamma '$       monotonicity

4. $\gamma \rightarrow \bot$       Ax4 and M.P. with 3

5. $\psi \rightarrow \gamma$       assumption

6. $\gamma$       M.P. 2, 5

7. $\bot$       M.P. 4, 6

Now applying Deduction theorem we get $\Gamma \; \vdash \; (\psi \rightarrow \gamma ) \rightarrow \bot$.

Again for proving $\Gamma \; \vdash \; \lnot (\psi \rightarrow \gamma )$ we do the following derivation:

$\Gamma \; \vdash$

1. $\psi '$       induction hypothesis

2. $\psi$       Ax3 and M.P. with 1

3. $\gamma '$       induction hypothesis

4. $\gamma \rightarrow \bot$       Ax4 and M.P. with 3

5. $\psi \wedge (\gamma \rightarrow \bot )$       Rule 1 on 2 and 4

6. $\psi \wedge (\gamma \rightarrow \bot ) \rightarrow \lnot (\psi \rightarrow \gamma )$       Axiom 14

7. $\lnot (\psi \rightarrow \gamma )$       M.P. 3, 4

Hence, again by Rule 1 it is derived $\Gamma \; \vdash \; \lnot (\psi \rightarrow \gamma ) \wedge [(\psi \rightarrow \gamma ) \rightarrow \bot ]$ i.e., $\Gamma \; \vdash \; \varphi '$.

Subcase 4.5: If $v(\gamma ) = 0$ and then $v(\varphi ) = 0$. Therefore by definition, $\gamma ' = \lnot \gamma \wedge (\gamma \rightarrow \bot )$, $\psi ' = \psi \wedge \lnot \psi$ and

$\varphi ' = \lnot \varphi \wedge (\varphi \rightarrow \bot ) = \lnot (\psi \rightarrow \gamma ) \wedge [(\psi \rightarrow \gamma ) \rightarrow \bot ]$. Since $\Gamma \; \vdash \; \psi '$ and $\Gamma \; \vdash \; \gamma '$, by Axiom 1, 2 and using M.P. we get

$$\Gamma \; \vdash \; \psi \text { and } \Gamma \; \vdash \; \gamma \rightarrow \bot .$$

Therefore, in this subcase $\Gamma \; \vdash \; \varphi '$ can be proved by following the same steps used in Subcase 4.4.

Hence, combining all the cases the Lemma 7.3.5 is proved. $\square$

Appendix B

Proof of the theorem 7.4.2:

Lemma. First, we show that by adding any theorem of $\mathrm {CPL}$ (Classical Propositional Logic) which is not a theorem of $\mathbb {L}\mathrm {PS}_3$, as an axiom schema in $\mathbb {L}\mathrm {PS}_3$, all the theorems of $\mathrm {CPL}$ can be proved.

Let $\varphi (p_{i_1}, p_{i_2}, \ldots , p_{i_n})$ be a theorem of $\mathrm {CPL}$ but not a theorem of $\mathbb {L}\mathrm {PS}_3$, where $p_{i_1}, p_{i_2}, \ldots , p_{i_n}$ are the propositional variables. Hence, for any valuation v from the set of all formulas of $\mathbb {L}\mathrm {PS}_3$ to $\mathrm {PS}_3$ for which

$$v(\varphi (p_{i_1}, p_{i_2}, \ldots , p_{i_n})) = 0$$

there must exist some $p_{i_l}$, $1 \le l \le n$ such that . Now using this fact without loss of generality we may assume that for any given valuation v we have $v(\varphi (p_{i_1}, p_{i_2}, \ldots , p_{i_n})) = 0$ iff for all $l \in \{1, \ldots , n\}$. It is guaranteed by the following fact: Suppose a formula $\psi (p_{r_1}, p_{r_2}, \ldots p_{r_{t+1}})$ is such that for all $l \in \{ 1, \ldots , t\}$ but . We then replace the propositional variable $p_{r_{t+1}}$

• by $\lnot (p_{r_1} \rightarrow p_{r_1})$ if $v(p_{r_{t+1}}) = 0$

• by $(p_{r_1} \rightarrow p_{r_1})$ if $v(P_{r_{t+1}}) = 1$

in the formula $\psi (p_{r_1}, p_{r_2}, \ldots p_{r_{t+1}})$ and therefore after replacing, the formula will get the value 0 iff all its propositional variables take the value . In this way we always get such a formula $\varphi (p_{i_1}, p_{i_2}, \ldots , p_{i_n})$.

Let us now assume $\sigma (p_{k_1}, p_{k_2}, \ldots , p_{k_m})$ be another arbitrarily chosen theorem of $\mathrm {CPL}$ which is not a theorem of $\mathbb {L}\mathrm {PS}_3$, where $p_{k_1}, p_{k_2}, \ldots , p_{k_m}$ are the propositional variables. It will be proved that $\sigma (p_{k_1}, p_{k_2}, \ldots , p_{k_m})$ can be derived from the axiom system of $\mathbb {L}\mathrm {PS}_3$ if it is extended by the new axiom schema $\varphi (p_{i_1}, p_{i_2}, \ldots , p_{i_n})$. Let $\varphi (p_{k_j})$ be the formula replacing each propositional variable of $\varphi (p_{i_1}, p_{i_2}, \ldots , p_{i_n})$ by $p_{k_j}$, for all $j \in \{1, \ldots , m\}$.

Claim: $\varphi (p_{k_1}) \wedge \varphi (p_{k_2}) \wedge \cdots \wedge \varphi (p_{k_m}) \rightarrow \sigma (p_{k_1}, p_{k_2}, \ldots , p_{k_m})$ is a theorem of $\mathbb {L}\mathrm {PS}_3$.

Proof of the claim. Let v be any valuation. Two cases could happen: either $v(\sigma (p_{k_1}, p_{k_2}, \ldots , p_{k_m})) = 0$ or $v(\sigma (p_{k_1}, p_{k_2}, \ldots , p_{k_m})) \ne 0$.

If $v(\sigma (p_{k_1}, p_{k_2}, \ldots , p_{k_m})) = 0$ and since $\sigma (p_{k_1}, p_{k_2}, \ldots , p_{k_m})$ is a theorem of $\mathrm {CPL}$ there must exist $p_{k_j}$ such that for some $j \in \{1, \ldots , m\}$. Hence, $v(\varphi (p_{k_j})) = 0$ and therefore

$$v(\varphi (p_{k_1}) \wedge \varphi (p_{k_2}) \wedge \cdots \wedge \varphi (p_{k_m}) \rightarrow \sigma (p_{k_1}, p_{k_2}, \ldots , p_{k_m})) = 1.$$

Again if $v(\sigma (p_{k_1}, p_{k_2}, \ldots , p_{k_m})) \ne 0$ then by the truth tables of $\mathrm {PS}_3$

$$v(\varphi (p_{k_1}) \wedge \varphi (p_{k_2}) \wedge \cdots \wedge \varphi (p_{k_m}) \rightarrow \sigma (p_{k_1}, p_{k_2}, \ldots , p_{k_m})) = 1.$$

Hence, for any valuation the formula

$$\begin{aligned} \varphi (p_{k_1}) \wedge \varphi (p_{k_2}) \wedge \cdots \wedge \varphi (p_{k_m}) \rightarrow \sigma (p_{k_1}, p_{k_2}, \ldots , p_{k_m}) \end{aligned}$$

always get the value 1. So by the completeness theorem of $\mathbb {L}\mathrm {PS}_3$ the claim is proved.

Now let us extend the axiom system of $\mathbb {L}\mathrm {PS}_3$ by including $\varphi (p_{i_1}, \ldots , p_{i_n})$ as an axiom schema. Let this system be denoted by $\mathbb {L}'\mathrm {PS}_3$. Then by the Rule 1, $\varphi (p_{k_1}) \wedge \varphi (p_{k_2}) \wedge \cdots \wedge \varphi (p_{k_m})$ is a theorem of $\mathbb {L}'\mathrm {PS}_3$. Now using M.P. we have $\sigma (p_{k_1}, p_{k_2}, \ldots , p_{k_m})$ as a theorem of the new system. Hence, the lemma is proved.

From this lemma it follows that for the enhanced system $\mathbb {L}'\mathrm {PS}_3$,

$$\begin{aligned} \Gamma \vdash _{\mathbb {L}'\mathrm {PS}_3} \varphi \text { iff } \Gamma \vdash _{\mathrm {CPL}} \varphi . \end{aligned}$$

$\square$