TU Wien:Einführung in wissensbasierte Systeme VU (Egly)/Übungen SS12/Blatt 2 - Beispiel 3

Consider the following arguments:

1. All humans are mortal. If someone is dead, he must be mortal. Aristotle is dead. Therefore, Aristotle must be human.
2. The Sphinx asked every person she met a riddle. If someone did not solve the riddle, she killed that person, but if someone knows the answer, the person is not killed. Oedipus has met the sphinx and was not killed. Therefore, he must have known the answer.

Translate the argument into predicate logic and show by TC1 that the argument is either correct or else extract an interpretation from the tableau showing that the argument is not correct.

Lösungsvorschlag

Nach üblichem Schema wollen wir zeigen:

$\phi \models \psi$

Daher nehmen wir für den indirekten Tableau Beweis an:

$\phi \land \neg \psi$

1

${\begin{aligned}\phi &=\forall x[H(x)\rightarrow M(x)]\land \forall x[D(x)\rightarrow M(x)]\\\psi &=D(a)\rightarrow H(a)\end{aligned}}$

${\begin{aligned}{\text{nnf}}(\phi )&=\forall x[\neg H(x)\lor M(x)]\land \forall x[\neg D(x)\lor M(x)]\\{\text{nnf}}(\neg \psi )&=D(a)\land \neg H(a)\end{aligned}}$

1: $\forall x[\neg H(x)\lor M(x)]$
2: $\forall x[\neg D(x)\lor M(x)]$
3: $D(a)\,$
4: $\neg H(a)\,$
5 (aus 1): $\neg H(a)\lor M(a)$
6 (aus 2): $\neg D(a)\lor M(a)$
7 (aus 6): $\neg D(a)$	8 (aus 6): $M(a)\,$
* mit 3	9: (aus 5) $\neg H(a)\,$	10 (aus 5): $M(a)\,$
	$\Uparrow$	$\Uparrow$

Wir haben also ein Gegenmodell $I\nvDash \psi$ für das Entailment gefunden:

${\begin{aligned}I(M(a))&={\text{t}}\\I(D(a))&={\text{t}}\\I(H(a))&={\text{f}}\end{aligned}}$

2

$M(x)\,$	met the sphinx
$R(x)\,$	was asked a riddle
$S(x)\,$	solved the riddle (knows the answer)
$K(x)\,$	person is killed

${\begin{aligned}\phi _{1}&=\forall x[M(x)\rightarrow R(x)]\\\phi _{2}&=\forall x[(R(x)\land \neg S(x))\rightarrow K(x)]\\\psi &=(M(a)\land \neg K(a))\rightarrow S(a)\end{aligned}}$

${\begin{aligned}{\text{nnf}}(\phi _{1})&=\forall x[\neg M(x)\lor R(x)]\\{\text{nnf}}(\phi _{2})&=\forall x[\neg R(x)\lor S(x)\lor K(x)]\\{\text{nnf}}(\neg \psi )&=M(a)\land \neg K(a)\land \neg S(a)\end{aligned}}$

1: $\forall x[\neg M(x)\lor R(x)]$
2: $\forall x[\neg R(x)\lor S(x)\lor K(x)]$
3: $M(a)\,$
4: $\neg K(a)\,$
5: $\neg S(a)\,$
6 (aus 2): $\neg R(a)\lor S(a)\lor K(a)$
7 (aus 1): $\neg M(a)\lor R(a)$
8 (aus 7): $\neg M(a)\,$	9 (aus 7): $R(a)\,$
* mit 3	10 (aus 6): $\neg R(a)\,$	11: $S(a)\,$	12: $K(a)\,$
	* mit 9	* mit 5	* mit 4