# Exercise Sheet 2 for: Adaptive Logics applied to the Philosophy of Science

Back to the main course site \(\def\impl{\supset} \def\Dab{{\sf Dab}}\) \(\newcommand{\vd}[1]{\vdash_{\bf #1}}\)

## Task 1

Let \(\Gamma = \{!p_1 \vee {!}p_2, !p_1 \supset {!}p_3, !p_2 \supset {!}p_4, !p_1 \vee q, !p_4 \vee q\}\).

- Is \(q\) a consequence with the reliability strategy?
- Is \(q\) a consequence with the minimal abnormality strategy?

## Task 2

Show that the following holds:

\(\phi \in \Phi(\Gamma)\) iff \(\phi\) is a choice set of \(\Sigma(\Gamma)\) and for all \(!A \in \phi\) there is a \(\Delta_A \in \Sigma(\Gamma)\) for which \(\{!A\} = \Delta_A \cap \phi\).

## Task 3

(This is a more difficult task: if you cannot solve it, no problem and we simply go through it together next time.)

Let \(\phi\) be a choice set of \(\Sigma(\Gamma)\). Show that there is a \(\psi \subseteq \phi\) such that \(\psi \in \Phi(\Gamma)\).

*Tip:* Let \(\phi = \{A_1, A_2, \ldots \}\). Use Task 1 to iteratively/recursively construct \(\psi\). For this, let \(\psi_0 = \phi\) and let \(\psi_{i+1}\) be the result of manipulating \(\psi_i\) in a way that is inspired by the result in Task 1. Then let \(\psi = \bigcap_{i \ge 1} \psi_i\). Now you show that \(\psi\) is a choice set and that it satisfies the property stated in Task 1.

## Task 4

Show that the following holds:

\(\Gamma \vd{CL_\circ} \Dab(\Delta)\) iff \(\Gamma \vd{CL_\circ^r} \Dab(\Delta)\).

## Task 5

Indicate mistakes in the following proof fragment from \(\Gamma = \{(\circ A \wedge \neg A) \vee (\circ B \wedge \neg B), (\circ A \wedge \neg A) \vee \neg (\circ B \wedge \neg B), \circ A, \circ B\}\):

1 | \((\circ A \wedge \neg A) \vee (\circ B \wedge \neg B)\) | PREM | \(\emptyset\) |

2 | \((\circ A \wedge \neg A) \vee \neg (\circ B \wedge \neg B)\) | PREM | \(\emptyset\) |

3 | \((\circ A \wedge \neg A)\) | 1,2; RU | \(\emptyset\) |

4 | \(\circ A\) | PREM | \(\emptyset\) |

5 | \(A\) | 4; RU | \(\{\circ A \wedge \neg A\}\) |

6 | \(\circ B\) | RU | \(\emptyset\) |

\(^\checkmark\) 7 | \(B\) | 6; RC | \(\{\circ B \wedge \neg B\}\) |