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Problem 4 - Predicate Logic

1.
Transform the following logical formula into clausal forms:

$\displaystyle \forall x \forall y P(x, y) \rightarrow Q(x, y) $

$\displaystyle \neg \forall x \exists y P(x, y) \rightarrow Q(x, y) $

2.
Using linear resolution , show that the following set of clauses is unsatisfiable:

$\displaystyle \{A, B\}, \{B, C\}, \{C, D\}, \{\neg C, \neg A\}, \{\neg D, \neg B\}, \{\neg B, \neg C\}$

Prenex form: $ \forall x \forall y P(x, y) \rightarrow Q(x, y) $.

Skolemized form: $ \forall x \forall y P(x, y) \rightarrow Q(x, y) $.

Without universal quantifiers: $ P(x, y) \rightarrow Q(x, y) $.

Adequate connectives: $ \neg P(x, y) \vee Q(x, y) $.

Conjunctive normal form (or clausal form): $ \neg P(x, y) \vee Q(x, y) $.


Prenex form: $ \exists x \forall y \neg (P(x, y) \rightarrow Q(x, y)) $.

Skolemized form: $ \forall y \neg (P(c_1, y) \rightarrow Q(c_1, y)) $.

Without universal quantifiers: $ \neg (P(c_1, y) \rightarrow Q(c_1, y)) $.

Adequate connectives: $ \neg (\neg P(c_1, y) \vee Q(c_1, y)) = P(c_1, y) \wedge \neg Q(c_1, y) $.

Truth table:
$ \neg$ $ P(c_1, y)$ $ \wedge$ $ \neg$ $ Q(c_1, y)$
T T F F T
F T T T F
T F F F T
T F F T F

Disjonctive normal form:

$ \neg \left((P(c_1, y) \wedge Q(c_1, y))
\vee (\neg P(c_1, y) \wedge Q(c_1, y))
\vee (\neg P(c_1, y) \wedge \neg Q(c_1, y))\right)$.

De Morgan's law yiels a conjunctive normal form (or clausal form):

$ (\neg P(c_1, y) \vee \neg Q(c_1, y))
\wedge (P(c_1, y) \vee \neg Q(c_1, y))
\wedge (P(c_1, y) \vee Q(c_1, y))$.


We use the following resolution rule:

$\displaystyle C_1 \cup \{l\}, C_2 \cup \{l^c\} \vdash C_1 \cup C_2$

where $ C_i$ is a clause and $ l^j$ a literal. The wffs are considered to be written in CNF using the standard set notation. $ C_i$ is refered to as a parent clause and $ C_1 \cup C_2$ is refered to as the resolvent.

In a linear resolution, we follow the standard resolution procedure but we take care of the fact that the previous resolvent is always a parent clause of the next resolution.

\includegraphics{linear-resolution.ps}

We yield the empty clause, that means that we set of clauses is not satisfiable, i.e., there is no interpretation thanks to which this set can be given a true value.


next up previous
Next: Problem 5 - Newton-Raphson Up: Information Science I Previous: Problem 3 - Digital
Reynald AFFELDT
2000-06-08