RUB | Fakultät für Philosophie und Erziehungswissenschaft | Institut für Philosophie II
    Workgroup Home | Christian tiny_christian.stra_er.jpg | Dunja tiny_dunja._e_elja.jpg | Jesse tiny_jesse.heyninck.jpg | Mathieu tiny_mathieu.beirlaen.jpg | FAQ for members

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

  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.

Author: Christian Straßer

Created: 2015-06-09 Tue 00:50

Emacs 24.5.1 (Org mode 8.2.10)

Validate