# Inconsistency-Adaptive Logics

## Table of Contents

- 1. Inconsistency-Adaptive Logics
- 2. Prioritising Adaptive Logics
- 2.1. Some abnormalities count more …
- 2.2. Semantics based on this idea
- 2.3. An alternative semantics
- 2.4. The two semantics are identical
- 2.5. A simple example
- 2.6. How to prove things?
- 2.7. Marking
- 2.8. Back to our example
- 2.9. Is there a problem with strong reassurance?
- 2.10. How does this help with reassurance? Abnormal parts and choice sets

## 1 Inconsistency-Adaptive Logics

### 1.1 Basic Motivation

- reasoning on the basis of inconsistent information
- use a paraconsistent logic
- however, they are weak, if monotonic
- why?
- disjunctive syllogism (If \(A\) and \({\sim} A \vee B\) then \(B\)) doesn't hold
- do you see why?

### 1.2 What to do?

- strengthen paraconsistent logics adaptively
- if we assume \(A \wedge {\sim}A\) not to hold we get that
- \(A\) and \({\sim}A \vee B\) implies \(B\)
- \({\sim}A\) and \(A \vee B\) implies \(B\)

### 1.3 So …

- we need to first settle for a paraconsistent lower limit logic
- a good candidate for the logical form of abnormalities is \({\sim}A \wedge A\)

### 1.4 Enters: CLuN

CLuN is defined by Modus Ponens (MP) and the following axiom schemata:

### 1.5 The semantics of CLuN

### 1.6 CLuN is weak …

no double negation elimination/introduction, no De Morgan Rules:

### 1.7 Now, let's go adaptive

Note that:

- \(A, {\sim} A \vee B \vdash_{CLuN} B \vee (A \wedge {\sim} A)\)
- \(A \vdash_{CLuN} {\sim}{\sim} A \vee (A \wedge {\sim}A)\)
- \({\sim}{\sim} A \vdash_{CLuN} A \vee ({\sim} A \wedge {\sim}{\sim} A)\)
- \({\sim}(A \vee B) \vdash_{CLuN} ({\sim} A \wedge {\sim} B) \vee ((A \vee B) \wedge {\sim}(A \vee B))\)
- etc.

Thus:

- \(A, {\sim} A \vee B \vdash_{CLuN^{x}} B\)
- \(A \vdash_{CLuN^{x}} {\sim}{\sim} A\)
- \({\sim}{\sim} A \vdash_{CLuN^{x}} A\)
- \({\sim}(A \vee B) \vdash_{CLuN^{x}} ({\sim} A \wedge {\sim} B)\)
- etc.

### 1.8 An example

Take \(\Gamma = \{ p, {\sim}p \vee q, r, {\sim}r\}\).

- What can you derive?

### 1.9 What about strengthening CLuN? Enters: CLuNs

{CLuNs} is {CLuN} plus De Morgan and plus double negation introduction and elimination

### 1.10 The (2-valued) semantics of CLuNs

### 1.11 The Flip/Flop problem

- suppose we use the same unrestricted form of abnormalities \(A \wedge {\sim}A\) as for CLuN
- Do you see a problem if we use this logic and reason, for instance, with \(\Gamma = \{ p, {\sim}p \vee q, r, {\sim}r\}\)?

### 1.12 What's the cure?

- Restrict the form of abnormalities to atoms!

## 2 Prioritising Adaptive Logics

### 2.1 Some abnormalities count more …

- Idea: structure the set of abnormalities!
- Instead of \(\Omega\) we use \(\langle\Omega_1, \Omega_2, \ldots \rangle\) where each \(\Omega_i\) is specified by a logical form

### 2.2 Semantics based on this idea

- start with the set \(\mathcal{M}_{\bf LLL}(\Gamma)\) of all LLL-models of \(\Gamma\)
- now select all \(\Omega_1\)-minimally abnormal models out of these: \(\mathcal{M}_1(\Gamma)\)
- then select all \(\Omega_2\)-minimally abnormal models out of \(\mathcal{M}_1(\Gamma)\) denoted by \(\mathcal{M}_2(\Gamma)\)
- go on like this
- the set of all selected models \(\mathcal{M}_p(\Gamma)\) is given by \(\bigcap_{i\ge 1}\mathcal{M}_i(\Gamma)\).
- the consequence relation is then defined as usual: \(\Gamma \Vdash A\) iff for all \(M \in \mathcal{M}_p(\Gamma)\), \(M \models A\)

### 2.3 An alternative semantics

Where \(\Delta \subseteq \bigcup_{i\ge 1} \Omega_i\) we define \(\Delta_i = \Delta \cap \Omega_i\).

- Lexicographic Ordering
\(\Delta \prec_{\rm lex} \Theta\) iff there is a \(i \ge 1\) such that

- for all \(j \le i\): \(\Delta_j = \Theta_j\)
- \(\Delta_i \subset \Theta_i\).

- Order Models according to abnormal parts and lex-order
Let \(M \sqsubset M'\) iff \(\Ab(M) \prec \Ab(M')\).

\(\Gamma \Vdash A\) iff for all \(\sqsubset\)-minimal models \(M\) in \(\mathcal{M}(\Gamma)\), \(M \models A\).

### 2.4 The two semantics are identical

We have to show that \(M \in \mathcal{M}_p(\Gamma)\) iff \(M\) is \(\sqsubset\)-minimal in \(\mathcal{M}(\Gamma)\).

- (\(\Rightarrow\))
- Suppose \(M \in \mathcal{M}_p(\Gamma)\).
- Assume for a contradiction that there is a \(M' \in \mathcal{M}(\Gamma)\) for which \(\Ab(M') \sqsubset \Ab(M)\).
- Hence, there is a minimal \(i\) for which \(\Ab(M')_i \subset \Ab(M)_i\) and \(\Ab(M')_j = \Ab(M)_j\) for all \(j < i\).
- Hence, both \(M, M' \in \mathcal{M}_j(\Gamma)\) for all \(j < i\).
- However, \(\Ab(M')_i \subset \Ab(M)_i\) contradicts the fact that \(M \in \mathcal{M}_i(\Gamma)\).

- (\(\Leftarrow\))
Similar (try it!).

### 2.5 A simple example

Take \(\Gamma = \{\circ_1 p, \circ_2 q, \circ_3r, \neg p \vee \neg q \vee \neg r\}\)

Here are some models

model | \(\circ_1 p \wedge \neg p\) | \(\circ_2q \wedge \neg q\) | \(\circ_3r \wedge \neg r\) |

\(M_1\) | 1 | 1 | 0 |

\(M_2\) | 1 | 0 | 0 |

\(M_3\) | 0 | 1 | 0 |

\(M_4\) | 0 | 1 | 1 |

\(M_5\) | 0 | 0 | 1 |

- Can you determine \(\mathcal{M}_i(\Gamma)\) for \(i \in \{1,2,3\}\)?
- What follows in the example?

### 2.6 How to prove things?

- basic idea: similar to minimal abnormality strategy
- just instead of \(\Phi_s(\Gamma)\) use a different set: \(\Phi_s^{\rm lex}(\Gamma)\) which we will define in a moment

### 2.7 Marking

A line \(l\) with an argument \(A\) and a condition \(\Delta\) is marked iff

- there is a \(\Theta \in \Phi_s^{\rm lex}(\Gamma)\) for which \(\Delta \cap \Theta = \emptyset\)
- for each \(\Theta \in \Phi_s^{\rm lex}(\Gamma)\) there is a line \(l'\) with argument \(A\) and a condition \(\Delta'\) for which \(\Delta' \cap \Theta = \emptyset\).

### 2.8 Back to our example

prove \(p \wedge q\) from \(\Gamma = \{\circ_1 p, \circ_2 q, \circ_3r, \neg p \vee \neg q \vee \neg r\}\)

### 2.9 Is there a problem with strong reassurance?

Let \(\Delta\) be a choice set of \(\Sigma(\Gamma)\). We show that there is a \(\prec\)-minimal choice set \(\Delta'\) of \(\Sigma(\Gamma)\) for which \(\Delta' \subseteq \Delta\).

- The construction
- Let \(\Sigma_1(\Gamma)\) be the set of all \(\Theta \subseteq \Omega_1\) such that \(\Dab(\Theta)\) is a minimal Dab-consequence of \(\Gamma\).
- We know that there \(\Delta_1\) is a choice set of \(\Sigma_1(\Gamma)\).
- We also know that there is a \(\Delta'_1 \subseteq \Delta_1\) that is a minimal choice set of \(\Sigma_1(\Gamma)\) (see one of the previous sessions).
- Let \(\Sigma_2(\Gamma)\) the set of all \(\Theta \setminus \Omega_1\) for which \(\emptyset \neq \Theta \setminus \Omega_1 \subseteq \Omega_2\), \(\Delta'_1 \cap \Theta = \emptyset\), and \(\Dab(\Theta)\) is a minimal Dab-consequence of \(\Gamma\).
- If \(\Delta'_1 \subset \Delta_1\) let \(\Delta'_2\) be any minimal choice set of \(\Sigma_2(\Gamma)\).
- Otherwise, \(\Delta_2\) is a choice set of \(\Sigma_2(\Gamma)\). Let \(\Delta'_2\) be a minimal choice set of \(\Sigma_2(\Gamma)\) such that \(\Delta'_2 \subseteq \Delta_2\).

- Let \(\Sigma_3(\Gamma)\) the set of all \(\Theta \setminus (\Omega_1 \cup \Omega_{2})\) for which \(\emptyset \neq \Theta \setminus (\Omega_1 \cup \Omega_2) \subseteq \Omega_3\), \((\Delta'_1 \cup \Delta'_{2}) \cap \Theta = \emptyset\), and \(\Dab(\Theta)\) is a minimal Dab-consequence of \(\Gamma\).
- If \(\Delta'_1 \cup \Delta_{2}' \subset \Delta_1 \cup \Delta_2\) let \(\Delta'_3\) be any minimal choice set of \(\Sigma_3(\Gamma)\).
- Otherwise, \(\Delta_3\) is a choice set of \(\Sigma_3(\Gamma)\). Let \(\Delta'_3\) be a minimal choice set of \(\Sigma_3(\Gamma)\) such that \(\Delta'_3 \subseteq \Delta_3\).

- We proceed similarly for all \(i \ge 4\).
- Let \(\Delta' = \bigcup_{i\ge 1} \Delta'_i\).

- Showing that we found the right \(\Delta'\)
- It is easy to see via induction that \(\Delta' \prec \Delta\) or \(\Delta' = \Delta\). (Try to prove this!)
- Assume there is a \(\Theta\) that is a choice set of \(\Sigma(\Gamma)\) and for which \(\Theta \prec \Delta'\).
- Thus, there is a minimal \(i\) for which \(\Theta_i \subset \Delta'_i\) and \(\Theta_j = \Delta'_j\) for all \(j < i\).
- Since \(\Delta'_i\) is a minimal choice set of \(\Sigma_i(\Gamma)\) there is a \(\Lambda \setminus \bigcup_{j < i} \Omega_j \in \Sigma_i(\Gamma)\) for which \(\Theta_i \cap (\Lambda \setminus \bigcup_{j < i} \Omega_j) = \emptyset\).
- Thus, \(\Theta_i \cap \bigcup_{j < i} \Omega_j \neq \emptyset\). However since \(\Delta'_i \cap \bigcup_{j < i} \Omega_j = \emptyset\) this is a contradiction.

- Let \(\Theta \in \Sigma(\Gamma)\) and let \(i\) be the highest index appearing in \(\Theta\).
- Suppose \(\Delta' \cap \bigcup_{j < i} \Theta_j = \emptyset\). Hence, \(\Theta \in \Sigma_i(\Gamma)\). Since \(\Delta'_i\) is a minimal choice set of \(\Sigma_i(\Gamma)\), \(\Delta' \cap \Theta \neq \emptyset\).
- This shows that \(\Delta'\) is a choice set of \(\Sigma(\Gamma)\).

- Altogether we have found a \(\Delta'\) that is a \(\prec\)-minimal choice set of \(\Sigma(\Gamma)\) and for which \(\Delta' \subseteq \Delta\).

### 2.10 How does this help with reassurance? Abnormal parts and choice sets

- from abnormal parts to choice sets
Where \(M \in \mathcal{M}(\Gamma)\), \(\Ab(M)\) is a choice set of \(\Sigma(\Gamma)\).

- from choice sets to abnormal parts
Where \(\Delta\) is a choice set of \(\Sigma(\Gamma)\) there is a model \(M \in \mathcal{M}(\Gamma)\) such that \(\Ab(M) \subseteq \Delta\).

- Proof
- Suppose there is no model \(M \in \mathcal{M}(\Gamma)\) for which \(\Ab(M) \subseteq \Delta\).
- Hence, \(\Gamma \cup (\Omega \setminus \Delta)^{\neg}\) is LLL-trivial.
- Hence, \(\Gamma \cup (\Omega \setminus \Delta)^{\neg} \Vdash A\) for any \(A\). Let \(A \in \Omega \setminus \Delta\)
- By the compactness of LLL there is a finite subset of \(\Theta^{\neg} \subseteq (\Omega \setminus \Delta)^{\neg}\) such that \(\Gamma \cup \Theta^{\neg} \Vdash A\).
- By the deduction theorem, \(\Gamma \Vdash \bigwedge\Theta^{\neg} \supset A\) and hence \(\Gamma \Vdash \Dab(\Theta\cup \{A\})\).
- Hence, there is a minimal \(\Theta' \subseteq \Theta \cup \{A\}\) such that \(\Gamma \Vdash \Dab(\Theta')\). (Note that \(\Theta' \in \Sigma(\Gamma)\).)
- Since \(\Delta\) is a choice set of \(\Sigma(\Gamma)\) and \(\Delta \cap \Theta = \emptyset\) this is a contradiction.

- Proof
- strong reassurance
Where \(M \in \mathcal{M}(\Gamma)\) there is a \(M' \in \mathcal{M}_p(\Gamma)\) for which \(\Ab(M') \subseteq \Ab(M)\).

- Proof
- Let \(M \in \mathcal{M}(\Gamma)\).
- Hence, \(\Ab(M)\) is a choice set of \(\Sigma(\Gamma)\).
- Thus, there is a \(\Delta\) that is a \(\prec\)-minimal choice set of \(\Sigma(\Gamma)\) for which \(\Delta \subseteq \Ab(M)\).
- Hence, there is a \(M' \in \mathcal{M}(\Gamma)\) for which \(\Ab(M') \subseteq \Delta\).
- Assume \(\Ab(M') \subset \Delta\). Then \(\Ab(M') \prec \Delta\). Since \(\Ab(M')\) is a choice set of \(\Sigma(\Gamma)\) this is a contradiction. Hence, \(\Ab(M') = \Delta\).
- Hence, \(M' \in \mathcal{M}_p(\Gamma)\).

- Proof