Axiomatisations of the Genuine Three-Valued Paraconsistent Logics $\mathbf {L3A_G}$ and $\mathbf {L3B_G}$

Abstract

Genuine Paraconsistent logics $\mathbf {L3A}$ and $\mathbf {L3B}$ were defined in 2016 by Béziau et al, including only three logical connectives, namely, negation disjunction and conjunction. Afterwards in 2017 Hernández-Tello et al, provide implications for both logics and define the logics $\mathbf {L3A_G}$ and $\mathbf {L3B_G}$. In this work we continue the study of these logics, providing sound and complete Hilbert-type axiomatic systems for each logic. We prove among other properties that $\mathbf {L3A_G}$ and $\mathbf {L3B_G}$ satisfy a restricted version of the Substitution Theorem, and that both of them are maximal with respect to Classical Propositional Logic. To conclude we make some comparisons between $\mathbf {L3A_G}$ and $\mathbf {L3B_G}$ and among other logics, for instance ${\mathbf {Int}}$ and some ${\mathbf {LFI}}$s.

Appendices

Appendix A: Some Proofs in $\mathbf {L3A_G}$

1.1 A.1: Proof of Lemma 1

Proof

(a) :

${\varvec{\vdash \lnot \lnot \varphi \rightarrow \varphi }}$.

(b) :

If ${\varvec{\vdash \varphi \rightarrow \psi }}$ and ${\varvec{\vdash \psi \rightarrow \sigma }}$ then ${\varvec{\vdash \varphi \rightarrow \sigma }}$.

By DT we have $\varphi \vdash \psi$ and $\psi \vdash \sigma$, then by Cut $\varphi \vdash \sigma$, finally applying DT again we get $\vdash \varphi \rightarrow \sigma$.

(c) :

${\varvec{ \vdash \big (\varphi \rightarrow (\psi \rightarrow \sigma ) \big ) \rightarrow \big (\psi \rightarrow (\varphi \rightarrow \sigma ) \big ) }}$.

(d) :

${\varvec{ \vdash (\lnot \lnot \psi \rightarrow \lnot \lnot \varphi ) \rightarrow (\lnot \varphi \rightarrow \lnot \psi )}}$.

(e) :

${\varvec{\vdash (\lnot \psi \rightarrow \lnot \varphi ) \rightarrow (\lnot \lnot \varphi \rightarrow \lnot \lnot \psi )}}$.

(f) :

${\varvec{\vdash \lnot \varphi \rightarrow \lnot \lnot \lnot \varphi }}$.

(g) :

${\varvec{\vdash (\lnot \varphi \wedge \lnot \lnot \varphi ) \rightarrow \psi }}$.

From $\lnot \varphi \wedge \lnot \lnot \varphi$ as hypothesis and $\lnot \lnot \varphi \rightarrow ( \lnot \varphi \rightarrow \psi )$, which is valid by WE. We obtain $\lnot \varphi \wedge \lnot \lnot \varphi \vdash \psi$, then applying DT we get the claim.

(h) :

${\varvec{ G(\varphi ), \varphi \vdash \psi }}$.

(i) :

${\varvec{ G(\varphi )\vdash \lnot \varphi }}$.

(j) :

${\varvec{ \vdash \big ( \varphi \rightarrow \psi \big ) \rightarrow \big ( G(\psi ) \rightarrow G(\varphi ) \big ) }}$.

(k) :

If ${\varvec{\varphi \vdash \psi }}$ and ${\varvec{ \sigma \vdash \xi }}$ then ${\varvec{ \varphi \wedge \sigma \vdash \psi \wedge \xi }}$.

(l) :

If ${\varvec{G(\varphi )\vdash \psi , N(\varphi )\vdash \psi }}$ and ${\varvec{ \lnot \lnot \varphi \vdash \psi }}$, then ${\varvec{ \vdash \psi }}$.

(m) :

${\varvec{\varphi \wedge \psi \vdash \sigma }}$ if and only if ${\varvec{\varphi , \psi \vdash \sigma }}$.

$\square$

1.2 A.2: Proof of Lemma 2

Proof

N1 :

${\varvec{\vdash G(\varphi ) \rightarrow \lnot \lnot (\lnot \varphi )}}.$

N2 :

${\varvec{\vdash N(\varphi )\rightarrow \lnot \lnot (\lnot \varphi )}}.$

N3 :

${\varvec{\vdash \lnot \lnot \varphi \rightarrow G(\lnot \varphi )}}.$

I1 :

${\varvec{\vdash G(\varphi )\rightarrow \lnot \lnot (\varphi \rightarrow \psi )}}.$

I2 :

${\varvec{\vdash \lnot \lnot \psi \rightarrow \lnot \lnot (\varphi \rightarrow \psi )}}.$

I3 :

${\varvec{\vdash (N(\varphi )\wedge N(\psi ))\rightarrow \lnot \lnot (\varphi \rightarrow \psi )}}.$

I4 :

${\varvec{\vdash (\lnot \lnot \varphi \wedge N(\psi ))\rightarrow N(\varphi \rightarrow \psi )}}.$

I5 :

${\varvec{\vdash (N(\varphi )\wedge G(\psi ))\rightarrow G(\varphi \rightarrow \psi )}}.$

I6 :

${\varvec{\vdash (\lnot \lnot \varphi \wedge G(\psi ))\rightarrow G(\varphi \rightarrow \psi )}}.$

C1 :

${\varvec{ \vdash G(\varphi )\rightarrow G(\varphi \wedge \psi )}}.$

C2 :

${\varvec{\vdash (N(\varphi )\wedge N(\psi ))\rightarrow N(\varphi \wedge \psi )}}.$

C3 :

${\varvec{\vdash (N(\varphi )\wedge \lnot \lnot \psi )\rightarrow \lnot \lnot (\varphi \wedge \psi )}}.$

C4 :

${\varvec{\vdash (\lnot \lnot \varphi \wedge N(\psi ))\rightarrow \lnot \lnot (\varphi \wedge \psi )}}.$

The proof is analogous to the case of C3 but starting with $\mathbf {L3A5}$ instead of $\mathbf {L3A4}$.

C5 :

${\varvec{\vdash (\lnot \lnot \varphi \wedge \lnot \lnot \psi )\rightarrow \lnot \lnot (\varphi \wedge \psi )}}.$

D1 :

${\varvec{\vdash \lnot \lnot \varphi \rightarrow \lnot \lnot (\varphi \vee \psi )}}.$

D2 :

${\varvec{\vdash (G(\varphi )\wedge N(\psi ))\rightarrow N(\varphi \vee \psi )}}.$

D3 :

${\varvec{\vdash (G(\varphi )\wedge N(\psi ))\rightarrow N(\varphi \vee \psi )}}.$

The proof is analogous to the proof of D2 but using Pos6 instead of Pos7.

D4 :

${\varvec{\vdash (N(\varphi )\wedge N(\psi ))\rightarrow N(\varphi \vee \psi )}}.$

D5 :

${\varvec{\vdash (G(\varphi )\wedge G(\psi ))\rightarrow G(\varphi \vee \psi )}}.$

$\square$

Appendix B: Some Proofs in $\mathbf {L3B_G}$

1.1 B.1: Proof of Lemma 6

Proof

(a) :

${\varvec{\vdash \lnot \lnot \varphi \rightarrow \varphi }}.$

(b) :

If ${\varvec{\vdash \varphi \rightarrow \psi }}$ and ${\varvec{\vdash \psi \rightarrow \sigma }}$ then ${\varvec{\vdash \varphi \rightarrow \sigma }}$.

See proof of Lemma 1, part b).

(c) :

${\varvec{ \vdash \big (\varphi \rightarrow (\psi \rightarrow \sigma ) \big ) \rightarrow \big (\psi \rightarrow (\varphi \rightarrow \sigma ) \big ) }}$.

See proof of Lemma 1, part c).

(d) :

${\varvec{ \varphi \wedge \psi \vdash \sigma }}$ if and only if ${\varvec{ \varphi , \psi \vdash \sigma }}$.

See proof of Lemma 1, part m).

(e) :

${\varvec{\vdash \big ( \varphi \wedge \lnot (\varphi \wedge \varphi ) \big ) \rightarrow \psi }}$.

(f) :

${\varvec{\vdash \big (\lnot \lnot \varphi \wedge \lnot (\varphi \wedge \varphi ) \big ) \rightarrow \psi }}$.

(g) :

${\varvec{ G(\varphi ), \varphi \vdash \psi }}$.

(h) :

${\varvec{ G(\varphi )\vdash \lnot \varphi }}$.

(i) :

${\varvec{ \lnot G(\varphi )\vdash \varphi }}$.

(j) :

${\varvec{ \varphi \vdash G\big (G( \varphi )\big ) }}$.

(k) :

${\varvec{ G\big (G(\varphi )\big ) \vdash \varphi }}$.

(l) :

${\varvec{ \vdash \big ( \varphi \rightarrow \psi \big ) \rightarrow \big ( G(\psi ) \rightarrow G(\varphi ) \big ) }}$.

(m) :

If ${\varvec{ \varphi \vdash \psi }}$ and ${\varvec{ \sigma \vdash \xi }}$ then ${\varvec{ \varphi \wedge \sigma \vdash \psi \wedge \xi }}$.

See proof of Lemma 1, part k).

(n) :

If ${\varvec{G(\varphi )\vdash \psi }}$, ${\varvec{N(\varphi )\vdash \psi }}$ and ${\varvec{\lnot G(\varphi ) \wedge \lnot N(\varphi )\vdash \psi }}$, then ${\varvec{\vdash \psi }}$.

$\square$

1.2 B.2: Proof of Lemma 7

Proof

N’1 :

${\varvec{\vdash G(\varphi ) \rightarrow D(\lnot \varphi )}}$.

N’2 :

${\varvec{\vdash N(\varphi )\rightarrow N(\lnot \varphi )}}$.

N’3 :

${\varvec{\vdash D(\varphi ) \rightarrow G(\lnot \varphi )}}$.

I’1 :

${\varvec{\vdash G(\varphi )\rightarrow D(\varphi \rightarrow \psi )}}$.

I’2 :

${\varvec{\vdash D(\psi ) \rightarrow D(\varphi \rightarrow \psi )}}$.

I’3 :

${\varvec{\vdash (N(\varphi )\wedge N(\psi ))\rightarrow D(\varphi \rightarrow \psi )}}$.

I’4 :

${\varvec{\vdash (D(\varphi )\wedge N(\psi ))\rightarrow N(\varphi \rightarrow \psi )}}$.

I’5 :

${\varvec{\vdash (N(\varphi )\wedge G(\psi ))\rightarrow G(\varphi \rightarrow \psi )}}$.

I’6 :

${\varvec{\vdash (D(\varphi ) \wedge G(\psi )) \rightarrow G(\varphi \rightarrow \psi )}}$.

C’1 :

${\varvec{ \vdash G(\varphi )\rightarrow G(\varphi \wedge \psi )}}$.

C’2 :

${\varvec{\vdash (N(\varphi )\wedge N(\psi ))\rightarrow D(\varphi \wedge \psi )}}$.

C’3 :

${\varvec{\vdash (D(\varphi )\wedge D(\psi )) \rightarrow D(\varphi \wedge \psi )}}$.

C’4 :

${\varvec{\vdash (N(\varphi ) \wedge D(\psi )) \rightarrow N(\varphi \wedge \psi )}}$.

C’5 :

${\varvec{\vdash (D(\varphi ) \wedge N(\psi )) \rightarrow N(\varphi \wedge \psi )}}$.

The proof is analogous to the case of C’4 but starting with $\mathbf {L3B5}$ instead of $\mathbf {L3B4}$.

D’1 :

${\varvec{\vdash D(\varphi ) \rightarrow D(\varphi \vee \psi )}}$.

D’2 :

${\varvec{\vdash (G(\varphi )\wedge N(\psi ))\rightarrow N(\varphi \vee \psi )}}$.

D’3 :

${\varvec{\vdash (N(\varphi )\wedge G(\psi ))\rightarrow N(\varphi \vee \psi )}}$.

The proof is analogous to the proof of D’2 but using Pos6 instead of Pos7.

D’4 :

${\varvec{\vdash (N(\varphi ) \wedge N(\psi )) \rightarrow N(\varphi \vee \psi )}}$.

D’5 :

${\varvec{\vdash (G(\varphi )\wedge G(\psi ))\rightarrow G(\varphi \vee \psi )}}$.

$\square$

1.3 B.3: Proof of Lemma 8

Proof

Let $\varphi \in Form(\mathbf {L3B_G})$ and v be a three-valuation in $\mathbf {L3B_G}$. The proof will be done by induction on the complexity of $\varphi$.

Base Case: If $\varphi =p$, with p an atomic formula, then $Atoms(\varphi )_v = \varphi _v= p_v$, but we know that $\varphi _v\vdash \varphi _v$, therefore $Atoms(\varphi )_v\vdash \varphi _v.$

Let us suppose now that for any $\beta \in Form(\mathbf {L3B_G})$ with lesser complexity than those of $\varphi$ the claim is true, namely $Atoms(\beta )_v\vdash \beta _v.$ Now we proceed to analyse what happen with each connective in $\mathbf {L3B_G}$.

Case ${\varvec{\lnot }}$: Suppose that $\varphi =\lnot \psi$, we know that $Atoms(\varphi )=Atoms(\psi )$ and by inductive hypothesis we have $Atoms(\varphi )_v\vdash \psi _{v}$. Depending what is the valuation of $\varphi$ we have one of the following cases:

• 1. If $v(\varphi )=0$, then $v(\psi )=2$, $\psi _v=D(\psi )$, $\varphi _v=G(\varphi )=G(\lnot \psi )$ and $Atoms(\varphi )_v \vdash D(\psi )$. By N’3 we have $D(\psi ) \vdash G(\lnot \psi )$ then applying Cut we conclude $Atoms(\varphi )_v\vdash G(\lnot \psi )$.

• 2. If $v(\varphi )=1$, then $v(\psi )=1$, $\psi _v=N(\psi )$, $\varphi _v = N(\varphi ) = N (\lnot \psi )$ and $Atoms(\varphi )_v \vdash N(\psi )$. By N’2 we have $N(\psi ) \vdash N(\lnot \psi )$ then applying Cut we conclude $Atoms(\varphi )_v \vdash N(\lnot \psi )$.

• 3. If $v(\varphi )=2$, then $v(\psi )=0$ then $\psi _ v=G( \psi )$, $\varphi _v = D(\varphi ) = D(\lnot \psi )$ and $Atoms(\varphi )_v\vdash G(\psi )$ and by property N’1, $G(\psi ) \vdash D(\lnot \psi )$ therefore $Atoms(\varphi )_v \vdash D(\lnot \psi )$.

  No matter what be the value of $\varphi$, we conclude $Atoms(\varphi )_v\vdash \varphi _v$.

Case ${\varvec{\rightarrow }}$: Suppose that $\varphi =\psi \rightarrow \sigma$. By inductive hypothesis we know that $Atoms(\psi )_v \vdash \psi _{v}$ and $Atoms(\sigma )_v \vdash \sigma _{v}$. Since $Atoms(\varphi )_v= Atoms(\psi )_v \cup Atoms(\sigma )_v$ then by Mon we have $Atoms(\varphi )_v \vdash \psi _v$ and $Atoms(\varphi )_v \vdash \sigma _v$. Depending what is the valuation of $\varphi$ we must consider the following cases:

• 1. If $v(\varphi )=0$ we have two possible scenarios $v(\psi )=1$ and $v(\sigma )=0$ or $v(\psi )=2$ and $v(\sigma )=0$, in any case we have $\varphi _v=G(\varphi )=G(\psi \rightarrow \sigma )$.

  • If $v(\psi )=1$ and $v(\sigma )=0$ by Definition 8, $\psi _v=N(\psi )$, $\sigma _v=G(\sigma )$, $Atoms(\varphi )_v \vdash N(\psi )$, $Atoms(\varphi )_v \vdash G(\sigma )$, and by property I’5 we know that $\vdash (N(\psi )\wedge G(\sigma )) \rightarrow G(\psi \rightarrow \sigma )$ then applying R-AND and MP we obtain $Atoms(\varphi )_v\vdash G(\psi \rightarrow \sigma )$.

  • If $v(\psi )=2$ and $v(\sigma )=0$, then we have $\psi _v=D(\psi )$, $\sigma _v=G(\sigma )$, $Atoms(\varphi )_v \vdash D(\psi )$, $Atoms(\varphi )_v \vdash G(\sigma )$, besides by I’6 we know that $\vdash (D(\psi ) \wedge G(\sigma )) \rightarrow G(\psi \rightarrow \sigma )$ again as in the previous case, applying R-AND and MP we have $Atoms(\varphi )_v\vdash G(\psi \rightarrow \sigma )$.

• 2. If $v(\varphi )=1$, then there is only one case, namely $v(\psi )=2$ and $v(\sigma )=1$. We have that $\psi _v= D(\psi )$, $\sigma _v = N(\sigma )$, $\varphi _v = N(\varphi ) = N(\psi \rightarrow \sigma )$, also we have that $Atoms(\varphi )_v \vdash D(\psi )$, $Atoms(\varphi )_v \vdash N(\sigma )$, and by property I’4 we know that $\vdash (D(\psi ) \wedge N(\sigma )) \rightarrow N(\psi \rightarrow \sigma )$ then applying R-AND and MP we obtain $Atoms(\varphi )_v\vdash N(\psi \rightarrow \sigma )$.

• 3. If $v(\varphi )=2$ then we have three possibilities, namely, $v(\psi )=0$ or $v(\sigma )=2$ or $v(\psi ) = v(\sigma ) = 1$, in any situation we have $\varphi _v=D(\varphi ) = D(\psi \rightarrow \sigma )$.

  • If $v(\psi )=0$ then $\psi _v=G(\psi )$, $Atoms(\varphi )_v \vdash G(\psi )$ and by property I’1 $G(\psi ) \vdash D(\psi \rightarrow \sigma )$ then by Cut we conclude that $Atoms(\varphi )_v \vdash D(\psi \rightarrow \sigma )$.

  • If $v(\sigma ) = 2$ then $\sigma _v= D(\sigma )$, $Atoms(\varphi )_v \vdash D(\sigma )$ and by I’2 we know that $D(\sigma ) \vdash D (\psi \rightarrow \sigma )$ therefore using Cut we derive $Atoms(\varphi )_v\vdash D(\psi \rightarrow \sigma )$.

  • If $v(\psi ) = v(\sigma ) = 1$ then $\psi _v = N(\psi )$, $\sigma _v = N(\sigma )$, $Atoms(\varphi )_v \vdash N(\psi )$, $Atoms(\varphi )_v \vdash N(\sigma )$ and by I’3 $N(\psi ) \wedge N(\sigma ) \vdash D(\psi \rightarrow \sigma )$ therefore using R-AND and Cut we obtain $Atoms(\varphi )_v\vdash D(\psi \rightarrow \sigma )$.

  Note that for each item, it was verified that: $Atoms(\varphi )_v\vdash \varphi _v$. So in this case the claim is verified.

Case ${\varvec{\wedge }}$: Suppose that $\varphi =\psi \wedge \sigma$. Analogously to the case of implication, we know that $Atoms(\varphi )_v \vdash \psi _v$ and $Atoms(\varphi )_v \vdash \sigma _v$. With base on the valuation of $\varphi$ we must study the following cases:

• 1. If $v(\varphi )=0$ we have two possible scenarios $v(\psi )=0$ or $v(\sigma )=0$, in any case we have $\varphi _v = G(\varphi ) = G(\psi \wedge \sigma )$.

  • If $v(\psi )=0$, then $\psi _v= G(\psi )$, $Atoms(\varphi )_v \vdash G(\psi )$, and by property C’1 we know that $G(\psi ) \vdash G(\psi \wedge \sigma )$ hence $Atoms(\varphi )_v \vdash G(\psi \wedge \sigma )$.

  • If $v(\sigma )=0$, using the same argument that in the previous case we have $Atoms(\varphi )_v \vdash G(\psi \wedge \sigma )$.

• 2. If $v(\varphi )=1$, then $v(\sigma )=1$ and $v(\psi )=2$ or $v(\sigma )=2$ and $v(\psi )=1$. We have that $\varphi _v= N(\varphi )=N(\sigma \wedge \psi )$.

  • If $v(\psi )=1, v(\sigma )=2$ then $\psi _v = N(\psi )$, $\sigma _v= D(\sigma )$, $Atoms(\varphi )_v \vdash N(\psi )$, $Atoms(\varphi )_v \vdash D(\sigma )$ and by property C’4 we have $N(\psi ) \wedge D(\sigma ) \vdash N(\psi \wedge \sigma )$ then using R-AND and Cut we conclude $Atoms(\varphi )_v \vdash N(\psi \wedge \sigma )$.

  • If $v(\psi )=2, v(\sigma )=1$ by a similar analysis to the previous case, now using C’5, we have $Atoms(\varphi )_v \vdash N(\psi \wedge \sigma )$.

• 3. If $v(\varphi )=2$ then we have two possible situations, $v(\psi )=2, v(\sigma )=2$ or $v(\psi )=1, v(\sigma )=1$, in any situation we have $\varphi _v = D(\varphi ) = D(\psi \wedge \sigma )$.

  • If $v(\psi )=2, v(\sigma )=2$ then $\psi _v = D(\psi )$, $\sigma _v= D(\sigma )$, $Atoms(\varphi )_v \vdash D(\psi )$, $Atoms(\varphi )_v \vdash D(\sigma )$ and by property C’3 $D(\psi ) \wedge D(\sigma ) \vdash D(\psi \wedge \sigma )$ then using R-AND and Cut we conclude $Atoms(\varphi )_v \vdash D(\psi \wedge \sigma )$.

  • If $v(\psi ) = v(\sigma ) = 1$ then $\psi _v = N(\psi )$, $\sigma _v = N(\sigma )$, $Atoms(\varphi )_v \vdash N(\psi )$, $Atoms(\varphi )_v \vdash N(\sigma )$ and by C’2 $N(\psi ) \wedge N(\sigma ) \vdash D(\psi \wedge \sigma )$ therefore using R-AND and Cut we obtain $Atoms(\varphi )_v\vdash D(\psi \wedge \sigma )$.

  In the three cases of conjunction we conclude that $Atoms(\varphi )_v\vdash \varphi _v$.

Case ${\varvec{\vee }}$: Suppose that $\varphi =\psi \vee \sigma$. Analogously to the previous case, we know that $Atoms(\varphi )_v \vdash \psi _v$ and $Atoms(\varphi )_v \vdash \sigma _v$. With base on the valuation of $\varphi$ we must study the following cases:

• 1. If $v(\varphi )=0$ then $v(\psi )=v(\sigma )=0$ from here $\psi _v = G(\psi )$, $\sigma _v= G(\sigma )$, $\varphi _v = G(\varphi ) = G(\psi \vee \sigma )$ and $Atoms(\varphi )_v \vdash G(\psi )$, $Atoms(\varphi )_v \vdash G(\sigma )$ and by property D’5, $G(\psi ) \wedge G(\sigma ) \vdash G( \psi \vee \sigma )$ then using R-AND and Cut we conclude $Atoms(\varphi )_v \vdash G(\psi \vee \sigma )$.

• 2. If $v(\varphi )=1$, we have three possible situations, $v(\psi )=1, v(\sigma )=0$ or $v(\psi )=0, v(\sigma )=1$ or $v(\psi ) = v(\sigma ) = 1$, in any situation we have $\varphi _v = N(\varphi ) = N(\psi \wedge \sigma )$.

  • If $v(\psi )=1, v(\sigma )=0$ then $\psi _v = N(\psi )$, $\sigma _v=G(\sigma )$, $Atoms(\varphi )_v \vdash N(\psi )$, $Atoms(\varphi )_v \vdash G(\sigma )$ and by property D’2 $N(\psi ) \wedge G(\sigma ) \vdash N(\psi \vee \sigma )$ then using R-AND and Cut we conclude $Atoms(\varphi )_v \vdash N(\psi \vee \sigma )$.

  • If $v(\psi )=0, v(\sigma )=1$ by a similar analysis to the previous case, using D’3 we have $Atoms(\varphi )_v \vdash N(\psi \vee \sigma )$.

  • If $v(\psi ) = v(\sigma ) = 1$ then $\psi _v = N( \psi$), $\sigma _v = N(\sigma )$, $Atoms(\varphi )_v \vdash N(\psi )$, $Atoms(\varphi )_v \vdash N(\sigma )$ and by D’4 $N(\psi ) \wedge N(\sigma ) \vdash N(\psi \vee \sigma )$ therefore using R-AND and Cut we obtain $Atoms(\varphi )_v\vdash N(\psi \vee \sigma )$.

• 3. If $v(\varphi )=2$ then we have two possible scenarios, $v(\psi )=2$ or $v(\sigma )=2$, in any situation we have $\varphi _v = D(\varphi ) = D(\psi \vee \sigma )$.

  • If $v(\psi )=2$, then $\psi _v= D(\psi )$, $Atoms(\varphi )_v \vdash D(\psi )$, and by property D’1 we know that $D(\psi ) \vdash D(\psi \vee \sigma )$ hence $Atoms(\varphi )_v \vdash D(\psi \vee \sigma )$.

  • If $v(\sigma )=2$, using the same argument that in the previous case we have $Atoms(\varphi )_v \vdash D(\psi \vee \sigma )$.

  Accordingly to the previous analysis the proof is complete.$\square$