4-Valued Semantics Under the OWA: A Deductive Database Approach

Abstract

In this paper, we introduce a novel approach for dealing with databases containing inconsistent information. Considering four-valued logics in the context of OWA (Open World Assumption), a database $\varDelta$ is a pair (E, R) where E is the extension and R the set of rules. In our formalism, the set E is a set of pairs of the form $\langle \varphi , \mathtt{v}\rangle$ where $\varphi$ is a fact and v is either t, or b, or f (meaning respectively true, inconsistent or false), given that unknown facts are not stored. Moreover the rules extend Datalog$^{neg}$ rules allowing their heads to be a negative atom.

We then define the notion of model of such a database, we show how to compute one particular model called semantics, and we investigate properties of this model. We also show how our approach applies to data integration and we review examples from the literature.

Appendices

A Proof of Lemma 1

Lemma 1

For every $\varDelta =(E,R)$ and all v-sets S, $S_1$ and $S_2$:

• 1. $\varSigma _{\varDelta }(S)$ is a v-set such that $E \preceq _k \varSigma _{\varDelta }(S)$.

• 2. If $S_1 \preceq _k S_2$, then $\varSigma _{\varDelta }(S_1) \preceq _k \varSigma _{\varDelta }(S_2)$.

Proof

• 1. $\varSigma _{\varDelta }(S)$ is a v-set because $\mathtt{v}_{\oplus }(\varphi )$ is unique, for a fixed $\varphi$. Moreover, every $\langle \varphi , \mathtt{v}\rangle$ in E also belongs to $\varGamma ^E_{\varDelta }(S)$, and as no rule can generate another pair involving $\varphi$, $\langle \varphi , \mathtt{v}\rangle$ is the only pair of $\varGamma ^E_{\varDelta }(S)$ involving $\varphi$. Hence $\langle \varphi , \mathtt{v}\rangle$ is in $\varSigma _{\varDelta }(S)$, and so $E \subseteq \varSigma _{\varDelta }(S)$ implying that $E \preceq _k \varSigma _{\varDelta }(S)$.

• 2. Let $v_1$ and $v_2$ denote respectively the valuations defined by $\varSigma _{\varDelta }(S_1)$ and $\varSigma _{\varDelta }(S_2)$, and let $\langle \varphi ,\mathtt{v}\rangle$ be in $\varSigma _{\varDelta }(S_1)$. If $\varphi$ occurs in E, then the previous point shows that $\langle \varphi ,\mathtt{v}\rangle$ also belongs to $\varSigma _{\varDelta }(S_2)$, and so $v_1(\varphi )=v_2(\varphi )$. Assuming now that $\varphi$ does not occur in E implies that rules whose head is $\varphi$ or $\lnot \varphi$ have generated pairs involving $\varphi$. We consider the following distinct cases:

  • Case 1: $v_1(\varphi )= \mathtt{t}$ or $v_1(\varphi )= \mathtt{f}$. Since in $S_1$, $\mathtt{v}_{\oplus }(\varphi )=\mathtt{t}$ (respectively $\mathtt{v}_{\oplus }(\varphi )=\mathtt{f}$), $\varGamma ^E_{\varDelta }(S_1)$ contains $\langle \varphi , \mathtt{t}\rangle$ (respectively $\langle \varphi , \mathtt{f}\rangle$) and not $\langle \varphi , \mathtt{b}\rangle$. Then, by Definition 4(1), inst(E, R) contains a rule $\rho$ such that $head(\rho )=\varphi$ (respectively $head(\rho )=\lnot \varphi$) and $v_{S_1}(body(\rho )) = \mathtt{t}$. Then, for every l in $body(\rho )$, if $l=\phi$, $v_{S_1}(\phi ) = \mathtt{t}$ and if $l=\lnot \phi$, $v_{S1}(\phi ) = \mathtt{f}$. Since $S_1 \preceq _k S_2$, for every $\phi$ occurring in $body(\rho )$, either $v_{S_2}(\phi )=v_{S_1}(\phi )$ or $v_{S_2}(\phi )=\mathtt{b}$. Hence, whatever the chosen implication ($\rightarrow$ or $\hookrightarrow$), $v_2(\varphi )=v_1(\varphi )$ or $v_2(\varphi )=\mathtt{b}$ and so, $v_1(\varphi ) \preceq _k v_2(\varphi )$.

  • Case 2: $v_1(\varphi )= \mathtt{b}$. Either $\langle \varphi , \mathtt{b}\rangle$ is in $\varGamma ^E_{\varDelta }(S_1)$ or not.

    • (i) If $\langle \varphi , \mathtt{b}\rangle$ is in $\varGamma ^E_{\varDelta }(S_1)$, then inst(E, R) contains $\rho$ such that $head(\rho )=\varphi$ or $head(\rho )=\lnot \varphi$, and $v_{S_1}(body(\rho )) = \mathtt{b}$. Hence, for every l in $body(\rho )$, if $l=\phi$, $v_{S_1}(\phi )$ is t or b and if $l=\lnot \phi$, $v_{S1}(\phi )$ is f or b, and at least one of these values is b. Since $S_1 \preceq _k S_2$, $v_{S_2}(\phi )$ is either t or b when $v_{S_1}(\phi )$ is t or b, and $v_{S_2}(\phi )$ is either f or b when $v_{S_1}(\phi )$ is f or b, and at least one of these values is b. Therefore $\langle \varphi ,\mathtt{b}\rangle$ belongs to $\varGamma ^E_{\varDelta }(S_2)$ and so, in $S_2$, $\mathtt{v}_{\oplus }(\varphi )=\mathtt{b}$. Thus, $v_1(\varphi ) \preceq _k v_2(\varphi )$.

    • (ii) If $\langle \varphi , \mathtt{b}\rangle$ is not in $\varGamma ^E_{\varDelta }(S_1)$, then $\langle \varphi , \mathtt{t}\rangle$ and $\langle \varphi , \mathtt{f}\rangle$ both belong to $\varGamma ^E_{\varDelta }(S_1)$, which implies that $\langle \varphi , \mathtt{b}\rangle$ is in $\varSigma _{\varDelta }(S_1)$. Hence inst(E, R) contains one rule whose head is $\varphi$ and one rule whose head is $\lnot \varphi$, and these rules apply for computing $\varGamma ^E_{\varDelta }(S_1)$ and for (i) above. Thus, in $S_2$ we have $\mathtt{v}_{\oplus }(\varphi )=\mathtt{b}$, showing that $v_1(\varphi ) \preceq _k v_2(\varphi )$.

      We have shown that for every $\varphi$ occurring $\varSigma _{\varDelta }(S_1)$, $v_1(\varphi ) \preceq _k v_2(\varphi )$. Now, for every $\varphi$ not occurring in $\varSigma _{\varDelta }(S_1)$, $v_1(\varphi )=\mathtt{n}$ which is the lowest truth value. Thus, $v_1(\varphi ) \preceq _k v_2(\varphi )$ holds, showing that $\varSigma _{\varDelta }(S_1) \preceq _k \varSigma _{\varDelta }(S_2)$ holds as well. Therefore, the proof is complete.    $\square$

B Proof of Proposition 1

Proposition 1

Given a database $\varDelta =(E,R)$, $\varSigma ^*_\varDelta$ is a minimal (with respect to set inclusion) model of $\varDelta$.

Proof

If $\varSigma ^*_{\varDelta }$ is not a model of $\varDelta$, then one rule $\rho$ of inst(E, R) is not valid in $\varSigma ^*_\varDelta$. In this case, $head(\rho )$ is not valid, while the conjunct defined by $body(\rho )$ is valid. Denoting $head(\rho )$ by $\varphi$ (respectively $\lnot \varphi$), either $v_{\varDelta } (\varphi )=\mathtt{n}$ or $v_{\varDelta }(\varphi )=\mathtt{f}$ (respectively $v_{\varDelta }(\varphi )=\mathtt{t}$). If $v_{\varDelta } (\varphi )=\mathtt{n}$, since $body(\rho )$ is valid in $\varSigma ^*_{\varDelta }$, we have $\varSigma _{\varDelta } (\varSigma ^*_{\varDelta } ) \ne \varSigma ^*_{\varDelta }$, which is not possible. Therefore, either $head(\rho )=\varphi$ and $v_{\varDelta }(\varphi )=\mathtt{f}$, or $head(\rho )=\lnot \varphi$ and $v_{\varDelta }(\varphi )=\mathtt{t}$, which is not possible by Definition 4. Thus, $\varSigma ^*_{\varDelta }$ is a model of $\varDelta$.

To show that $\varSigma ^*_\varDelta$ is a minimal model, let $\sigma$ be a nonempty subset of $\varSigma ^*_\varDelta$, and assume that $S=\varSigma ^*_\varDelta {\setminus } \sigma$ is a model of $\varDelta$. Let k be the least integer such that $\varSigma ^{k-1} \cap \sigma =\emptyset$ and $\varSigma ^{k} \cap \sigma \ne \emptyset$. We notice that k exists such that $k>0$ because, since S is a model of $\varDelta$, it holds that $E \subseteq S$ and so, since $\varSigma ^0=E$, we have $\varSigma ^{0} \cap \sigma =\emptyset$. Now, let $\langle \varphi , \mathtt{v}\rangle$ be in $\varSigma ^{k} \cap \sigma$ but not in $\varSigma ^{k-1}$. In this case, $v_S(\varphi )=\mathtt{n}$ and as above, there exists one rule $\rho$ in inst(E, R) such that $head(\rho )$ is either $\varphi$ or $\lnot \varphi$ and in $\varSigma ^{k-1}$, $head(\rho )$ is not valid, while the conjunct defined by $body(\rho )$ is valid. Since $\varSigma ^{k-1}\subseteq S$, $body(\rho )$ is valid in S while $head(\rho )$ is not. This is a contradiction and so, the proof is complete.   $\square$

C Proof of Proposition 2

Proposition 2

Let $\varDelta =(E,R)$ be such that for every rule $\rho$ in R, $head(\rho )$ is a positive literal. For all minimal models $M_1$ and $M_2$ of $\varDelta$, the following holds: (i) $\mathsf{V}(M_1) = \mathsf{V}(M_2)$ and (ii) $\mathsf{F}(M_1) = \mathsf{F}(M_2)$.

Proof

The proposition is a consequence of Lemma 2 shown next.   $\square$

Lemma 2

Let $\varDelta =(E,R)$ be such that for every rule $\rho$ in R, $head(\rho )$ is a positive literal. For every minimal model M of $\varDelta$, the following holds:

• 1. $\mathsf{F}(\varSigma ^*_\varDelta ) = \mathsf{F}(M)$.

• 2. $\mathsf{V}(\varSigma ^*_\varDelta )=\mathsf{V}(M)$.

Proof

• 1. As computing $\varSigma ^*_\varDelta$ starts from E and generates no other false facts, $\mathsf{F}(\varSigma ^*_\varDelta ) = \mathsf{F}(E)$. Since $\mathsf{F}(E) \subseteq \mathsf{F}(M)$, we obtain $\mathsf{F}(\varSigma ^*_\varDelta ) \subseteq \mathsf{F}(M)$.

  Assuming that $\mathsf{F}(M) \not \subseteq \mathsf{F}(\varSigma ^*_\varDelta )$, let $\varphi$ be in $\mathsf{F}(M) {\setminus } \mathsf{F}(\varSigma ^*_\varDelta )$. Denoting by $M'$ the set $M {\setminus } \{\langle \varphi , \mathtt{f}\rangle \}$, we show that $M'$ is a model of $\varDelta$ and thus that we obtain a contradiction since M is assumed to be minimal. To show that $M'$ is a model of $\varDelta$, we first note that $E \subseteq M'$ holds because so does $E \subseteq M$ and $\langle \varphi , \mathtt{f}\rangle$ is not E. Thus, assuming that $M'$ is not a model of $\varDelta$ entails that there exists $\rho$ in inst(E, R) that is not valid in $M'$. Hence, independently from the chosen implication $\rightarrow$ or $\hookrightarrow$, $body(\rho )$ is valid in $M'$ whereas $head(\rho )$ is not. However, since $body(\rho )$ is valid in $M'$, $\varphi$ does not occur in $body(\rho )$, implying that $body(\rho )$ is also valid in M. Hence, $head(\rho )$ must be valid in M and so, $head(\rho )=\varphi$. This is a contradiction with the fact that $\varphi$ is assumed to be in $\mathsf{F}(M)$.

• 2. We prove that $\mathsf{V}(\varSigma ^*_\varDelta ) \subseteq \mathsf{V}(M)$ by induction on k. Indeed, $\mathsf{V}(\varSigma ^0) \subseteq \mathsf{V}(M)$ holds because $E =\varSigma ^0$. Then, for $k>0$, assume that $\mathsf{V}(\varSigma ^{k-1}) \subseteq \mathsf{V}(M)$ and let $\varphi$ be in $\mathsf{V}(\varSigma ^k){\setminus } \mathsf{V}(M)$. Since $\mathsf{V}(E) \subseteq \mathsf{V}(M)$, $\varphi$ is not in $\mathsf{V}(E)$, $\varphi$ occurs in $\varSigma ^k$ due to a rule $\rho$. Thus, there exists $\rho$ in inst(E, R) such that $body(\rho )$ is valid in $\varSigma ^{k-1}$ and $head(\rho )=\varphi$. Since $\mathsf{V}(\varSigma ^{k-1}) \subseteq \mathsf{V}(M)$, $body(\rho )$ is valid in M and as $\rho$ must be valid in M, so is $\varphi$. We thus obtain a contradiction with the fact that $\varphi$ is assumed not to be in $\mathsf{V}(M)$. Hence $\mathsf{V}(\varSigma ^k)\subseteq \mathsf{V}(M)$ and thus, $\mathsf{V}(\varSigma ^*_\varDelta ) \subseteq \mathsf{V}(M)$. Now let $M'=\{\langle \varphi , \mathtt{v}\rangle \in M~|~\varphi \in \mathsf{V}(\varSigma ^*_\varDelta )\} \cup \{\langle \varphi , \mathtt{f}\rangle \in M~|~\varphi \in \mathsf{F}(\varSigma ^*_\varDelta )\}$. We notice that $M' \subseteq M$, and since $\mathsf{V}(\varSigma ^*_\varDelta ) \subseteq \mathsf{V}(M)$ and $\mathsf{F}(\varSigma ^*_\varDelta ) = \mathsf{F}(M)$, we have $\mathsf{V}(\varSigma ^*_\varDelta ) = \mathsf{V}(M')$ and $\mathsf{F}(\varSigma ^*_\varDelta ) = \mathsf{F}(M')$. We show that $M'$ is a model of $\varDelta$ and thus that $M'=M$ since M is assumed to be minimal. To show that $M'$ is a model of $\varDelta$, we first prove that $E \subseteq M'$. Indeed, as every $\langle \varphi , \mathtt{v}\rangle$ in E is also in $\varSigma ^*_\varDelta$ and in M, we have the following:

  • If $\mathtt{v} = \mathtt{f}$ then $\varphi$ is in $\mathsf{F}(E)$ thus in $\mathsf{F}(\varSigma ^*_\varDelta )$. In this case, $\langle \varphi , \mathtt{v}\rangle$ is in $M'$.

  • Otherwise, $\mathtt{v} = \mathtt{t}$ or $\mathtt{v} = \mathtt{b}$, that is $\varphi$ is in $\mathsf{V}(E)$. Thus $\varphi$ is in $\mathsf{V}(\varSigma ^*_\varDelta )$ and as $\langle \varphi , \mathtt{v}\rangle$ is in M, $\langle \varphi , \mathtt{v}\rangle$ is also in $M'$.

Every rule $\rho$ in inst(E, R) is valid in $M'$, because if $body(\rho )$ is valid in $M'$ then $body(\rho )$ is also valid in $\varSigma ^*_\varDelta$ and so, $head(\rho )$ is in $\mathsf{V}(\varSigma ^*_\varDelta )$. Thus $head(\rho )$ is in $\mathsf{V}(M')$. Hence, $M'=M$, showing that $\mathsf{V}(\varSigma ^*_\varDelta ) = \mathsf{V}(M)$.    $\square$