Übung 1 Löse folgendes Erfüllbarkeitsproblem mit dem

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Übung 1 Löse folgendes Erfüllbarkeitsproblem mit dem Davis-Putnam-Algorithmus! (rs)  (¬pq)  (¬rs)  (¬qp)  (¬r¬sp)  (¬p¬qr)  (¬p¬q¬r) p ¬p (rs)  (¬rs)  (¬q)  (¬r¬s) (rs)  (q)  (¬rs)  (¬qr)  (¬q¬r) q (unit) ¬q (unit) (rs)  (¬rs)  (r)  (¬r) (rs)  (¬rs)  (¬r¬s) s (pur literal) ¬r r (r)  (¬r) (s) (s)  (¬s) s (unit) Wid ok Wid

Übung 2 Konstruiere unter Verwendung des Stålmarck-Algorithmus eine Belegung, die jede der folgenden Formeln erfüllt! (p  q) , (p  r) , ((q r)  p) , (q  (s  t)) , (q  (s  u)) , (q  ((t  u)  s)) keine Simple rules anwendbar  Dilemma(p) Tripelmenge: 1  (p  q) 1  (p  r) x1  (q  r) 1  (x1  p) x2  (s  t) 1  (q  x2) x3  (s  u) 1  (q  x3) x4  (t  u) x5  (x4  s) 1  (q  x5) ¬p  1  (0  q) 1  (0  r) x1  (q  r) 1  (x1  0) x2  (s  t) 1  (q  x2) x3  (s  u) 1  (q  x3) x4  (t  u) x5  (x4  s) 1  (q  x5) ¬p + simple 1  (0  1) 1  (0  0) 0  (1  0) 1  (s  t) 1  (1  1) 1  (s  u) x4  (t  u) 1  (x4  s) p  1  (1  q) 1  (1  r) x1  (q  r) 1  (x1  1) x2  (s  t) 1  (q  x2) x3  (s  u) 1  (q  x3) x4  (t  u) x5  (x4  s) 1  (q  x5) p + simple  1  (1  1) 1  (s  t) 1  (s  u) x4  (t  u) 1  (x4  s)  q x2 x3 x5 x1=r p = r

Alle Tripel erfüllt,  erfüllende Belegungen q=1 t=1 p=r=0 s=u=0 Dilemma(s) Resultat nach Dilemma(p) 1  (p  1) 1  (p  r) r  (1  r) 1  (r  p) 1  (s  t) 1  (1  1) 1  (s  u) x4  (t  u) 1  (x4  s) ¬s  1  (p  1) 1  (p  p) p  (1  p) 1  (0  t) 1  (1  1) 1  (0  u) x4  (t  u) 1  (x4  0) ¬s + simple 1  (p  1) 1  (p  p) p  (1  p) 1  (0  1) 1  (1  1) 1  (0  0) 0  (1  0) s  1  (p  1) 1  (p  p) p  (1  p) 1  (1  t) 1  (1  1) 1  (1  u) x4  (t  u) 1  (x4  1) s + simple 1  (p  1) 1  (p  p) p  (1  p) 1  (1  1) q,x2,x3,x5,x1=r,p=r,t,x4=u,u=s Dilemma(t) Resultat nach Dilemma(s) 1  (p  1) 1  (p  p) p  (1  p) 1  (s  1) 1  (1  1) 1  (s  s) s  (1  s) Alle Tripel erfüllt,  erfüllende Belegungen q=1 t=1 p=r=0 s=u=0 q=1 t=1 p=r=0 s=u=1 q=1 t=1 p=r=1 s=u=0 q=1 t=1 p=r=1 s=u=1

Übung 3 Generiere die aussagenlogische Formel, die einen kreisfreien Pfad der Länge 2 für den 3-Bit-Zähler (Anfangszustand: alle Bits auf Null) beschreibt, der die Formel F (x1 U ¬ x0) bezeugt! ¬x0(0) ¬x1(0) ¬x2(0) (x0(1)   x0 (0) )(x1(1)  (¬ x1(0)  x0(0))) (x2(1)  (¬ x2(0) (x0(0)  x1(0) )))  (x0(2)   x0(1) )(x1(2)  (¬ x1(1)  x0(1))) (x2(2)  (¬ x2(1) (x0(1)  x1(1) )))  (¬x0(0)  [(¬x0(1)  (x1(0)  ¬x0(1))]  [(¬x0(2)  (x1(1)  ¬x0(2))  (x1(0)  x1(1)  ¬x0(2)) ]) Ist diese Formel erfüllbar? Ja. x0(0),x1(0),x2(0),x1(1),x2(1),x0(2),x2(2) = 0 x0(1),x1(2) = 1