# Some more meta-theory

$$\def\Dab{{\rm Dab}}$$

## 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 $$\mathcal{M}^m(\Gamma) = \mathcal{M}^m(\Gamma \cup \Delta)$$.

1. 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 $$\Delta \subseteq Cn^m(\Gamma)$$, $$Cn^m(\Gamma) = Cn^m(\Gamma \cup \Delta)$$.

### 2.6 Fixed Point Property

$$Cn^m(\Gamma) = Cn^m(Cn^m(\Gamma))$$.

1. Proof

Obviously $$Cn^m(\Gamma) \subseteq Cn^m(\Gamma)$$ and $$\Gamma \cup Cn^m(\Gamma) = Cn^m(\Gamma)$$. The rest follows by the corollary.

Created: 2015-06-09 Tue 00:50

Emacs 24.5.1 (Org mode 8.2.10)

Validate