### 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?

• 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:

### 1.6 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

### 1.11 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!

### 2.1 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
1. Example

$$\circ_i A \wedge \neg A$$ (rate the trustworthiness of your evidence: the lower $$i$$ the higher the trustworthiness of the evidence)

### 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 $$\Delta_i = \Delta \cap \Omega_i$$.

1. Lexicographic Ordering

$$\Delta \prec_{\rm lex} \Theta$$ iff there is a $$i \ge 1$$ such that

• for all $$j \le i$$: $$\Delta_j = \Theta_j$$
• $$\Delta_i \subset \Theta_i$$.
2. 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 $$\mathcal{M}(\Gamma)$$.

1. ($$\Rightarrow$$)
• 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 $$M \in \mathcal{M}_i(\Gamma)$$.
2. ($$\Leftarrow$$)

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
1. $$\Phi_s^{\rm lex}(\Gamma)$$

$$\Phi_s^{\rm lex}(\Gamma)$$ is the set of all choice sets of $$\Sigma(\Gamma)$$ that are $$\prec_{\rm lex}$$-minimal.

(Recall that $$\Sigma(\Gamma)$$ is the set of all $$\Theta \subseteq \Omega$$ such that $$\Dab(\Theta)$$ is a minimal Dab-formula at stage $$s$$.)

### 2.7 Marking

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

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

as before

### 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 $$\Delta' \subseteq \Delta$$.

1. 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 $$\Delta' = \bigcup_{i\ge 1} \Delta'_i$$.
2. 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

1. from abnormal parts to choice sets

Where $$M \in \mathcal{M}(\Gamma)$$, $$\Ab(M)$$ is a choice set of $$\Sigma(\Gamma)$$.

1. Proof

Let $$\Delta \in \Sigma(\Gamma)$$. Hence, $$\Gamma \Vdash_{\bf LLL} \Dab(\Delta)$$. Hence, $$M \models \Dab(\Delta)$$. Hence, there is a $$A \in \Delta$$ such that $$M \models A$$ and thus $$A \in \Ab(M)$$.

2. 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 $$\Ab(M) \subseteq \Delta$$.

1. 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.
3. strong reassurance

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

1. 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, $$M' \in \mathcal{M}_p(\Gamma)$$.

Created: 2015-07-07 Tue 03:01

Emacs 24.5.1 (Org mode 8.2.10)

Validate