• Keine Ergebnisse gefunden

Nonmonotonic Reasoning

N/A
N/A
Protected

Academic year: 2022

Aktie "Nonmonotonic Reasoning"

Copied!
1
0
0

Wird geladen.... (Jetzt Volltext ansehen)

Volltext

(1)

Faculty of Computer Science Institute of Theoretical Computer Science, Chair of Automata Theory

Nonmonotonic Reasoning

Winter Semester 2017/18

Exercise Sheet 1 18th October 2017

Dr. (habil.) Anni-Yasmin Turhan

Exercise 1.1 We consider substitutions. The compositionof two substitutions σ= [X1/t1, . . . ,Xn/tn]and ρ= [Y1/s1, . . . ,Yms/sm]is defined as follows

σρ=Xi/tiρ|Xi6=tiρfor1≤i≤n ∪Yj/sj|Yj 6∈ {Xi, . . . ,Xn}for1≤j≤m . (a) Given the substitutionsσ= [V1/p(), V2/p0(),V3/V4]andρ= [V2/V3, V4/f2(V3), V1/V4]together

with the term

t= f

g(V1),g2 p(),V2),f2(V4).

Is ϕ σρa ground substitution?

(b) Show that substitions are closed under composition.

Exercise 1.2 “For any formula ϕand admissible substitutionσ, the formula∀Xϕ−→ϕσ is a tautology.”

Does this claim hold or not? How can one show this?

Exercise 1.3 We are turning to Default logic. Devise a default theory that models the bike shop domain.

1

Referenzen

ÄHNLICHE DOKUMENTE

Faculty of Computer Science Institute of Theoretical Computer Science, Chair of Automata Theory.. Description Logic Winter

Faculty of Computer Science Institute of Theoretical Computer Science, Chair of Automata Theory.. Description Logic Winter

Faculty of Computer Science Institute of Theoretical Computer Science, Chair of Automata Theory. Nonmonotonic Reasoning Winter

Exercise 5.2 Devise a default theory with three extensions and compute all brave and all cautious consequences (modulo equivalence) of it. Exercise 5.3 Express some of the examples

Nonmonotonic Reasoning Winter Semester 2017/18 Exercise Sheet 9 – Inference relations 16th January

Gabriele R¨ oger (University of Basel) Theory of Computer Science April 26, 2021 5 /

Gabriele R¨ oger (University of Basel) Theory of Computer Science April 8, 2019 2 / 21?.

Prove with a direct proof: for all finite sets S, the power set P(S) has cardinality 2