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

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

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

1. Is $$q$$ a consequence with the reliability strategy?
2. Is $$q$$ a consequence with the minimal abnormality strategy?

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 $$\{!A\} = \Delta_A \cap \phi$$.

(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.

Show that the following holds:
$$\Gamma \vd{CL_\circ} \Dab(\Delta)$$ iff $$\Gamma \vd{CL_\circ^r} \Dab(\Delta)$$.

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 $$\{\circ B \wedge \neg B\}$$