# Some more meta-theory

## Table of Contents

\(\def\Dab{{\rm Dab}}\)

## 1 Deduction and Resolution Theorem

### 1.1 Recall

- Deduction Theorem: \(\Gamma \cup \{A\} \vdash B\) implies \(\Gamma \vdash A \supset B\).
- Resolution Theorem: \(\Gamma \vdash A \supset B\) implies \(\Gamma \cup \{A\} \vdash B\).

### 1.2 For reliability?

- What about the deduction theorem?
- Consider \(\Gamma = \{(\circ p \wedge \neg p) \vee (\circ q \wedge \neg q), \circ p, \circ q\}\).
- Then \(\Gamma \cup \{\neg(\circ p \wedge \neg p)\}\vdash^r p\) while \(\Gamma \nvdash^r \neg(\circ p \wedge \neg p) \supset p\).

- What about the resolution theorem?
- Consider \(\Gamma_r = \{\circ p, \neg(\circ q \wedge \neg q) \vee \neg p\}\).
- Then \(\Gamma_r \vdash^r p\) and hence \(\Gamma_r \vdash^r (\circ q \wedge \neg q) \supset p\)
- but \(\Gamma_r \cup \{\circ q \wedge \neg q\} \nvdash^r p\).

### 1.3 For minimal abnormality?

- What about the resolution theorem?
- Consider \(\Gamma_r\). Same argument as above.

- What about the deduction theorem?
- Suppose \(\Gamma \cup \{A\} \vdash^m B\).
- Let \(M\) be a minimally abnormal model of \(\Gamma\).
- Assume that \(M \not\models A \supset B\).
- Hence, \(M\) is not a minimally abnormal model of \(\Gamma \cup \{A\}\).
- Hence, \(M \models A\) and \(M \models \neg B\).
- Hence, \(M\) is a model of \(\Gamma \cup \{A\}\).
- By strong reassurance there is a minimally abnormal model \(M'\) of \(\Gamma \cup \{A\}\) such that \({\sf Ab}(M') \subset {\sf Ab}(M)\).
- However, since \(M'\) is a model of \(\Gamma\) this is a contradiction to the fact that \(M\) is a minimally abnormal model of \(\Gamma\).

## 2 Cautious Properties

### 2.1 Lemma

If \(\Gamma \Vdash^m A\) for all \(A \in \Delta\) then \(\mathcal{M}^m(\Gamma) = \mathcal{M}^m(\Gamma \cup \Delta)\).

- Proof
- Suppose \(\Gamma \Vdash^m A\) for all \(A \in \Delta\).
- Let \(M \in \mathcal{M}^m(\Gamma)\).
- By the supposition, \(M \in \mathcal{M}(\Gamma \cup \Delta)\).
- Suppose \(M' \in \mathcal{M}(\Gamma \cup \Delta)\) such that \({\sf Ab}(M') \subset {\sf Ab}(M)\).
- However, since \(M' \in \mathcal{M}(\Gamma)\) this is a contradiction to the fact that \(M \in \mathcal{M}^m(\Gamma)\).
- Thus, \(M \in \mathcal{M}^m(\Gamma \cup \Delta)\).
- Let now \(M \in \mathcal{M}^m(\Gamma \cup \Delta)\).
- Suppose there is a \(M' \in \mathcal{M}(\Gamma)\) such that \({\sf Ab}(M') \subseteq {\sf Ab}(M)\).
- By the main supposition, \(M' \in \mathcal{M}(\Gamma \cup \Delta)\).
- Thus, \({\sf Ab}(M) = {\sf Ab}(M')\) which shows that \(M \in \mathcal{M}^m(\Gamma)\).

### 2.2 Cautious Monotonicity

Where \(\Gamma \Vdash^m A\) for all \(A \in \Delta\), \(\Gamma \Vdash^m B\) implies \(\Gamma \cup \Delta \Vdash^m B\).

### 2.3 Cautious Transitivity

Where \(\Gamma \Vdash^m A\) for all \(A \in \Delta\), \(\Gamma \cup \Delta \Vdash^m B\) implies \(\Gamma \Vdash^m B\).

### 2.4 Proof

Both properties follow immediately with the lemma.

### 2.5 Corollary

Where \(Cn^m(\Gamma) = \{ A \mid \Gamma \Vdash^m A\}\) and \(\Delta \subseteq Cn^m(\Gamma)\), \(Cn^m(\Gamma) = Cn^m(\Gamma \cup \Delta)\).