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sav08:herbrand_s_expansion_theorem [2008/04/01 15:59] giuliano |
sav08:herbrand_s_expansion_theorem [2015/04/21 17:30] (current) |
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| Define | Define | ||
| - | \[ | + | \begin{equation*} |
| expand(C) = \{ subst(\{x_1 \mapsto t_1,\ldots,x_n \mapsto t_n\})(C) \mid FV(C) = \{x_1,\ldots,x_n\}\ \land\ t_1,\ldots,t_n \in GT \} | expand(C) = \{ subst(\{x_1 \mapsto t_1,\ldots,x_n \mapsto t_n\})(C) \mid FV(C) = \{x_1,\ldots,x_n\}\ \land\ t_1,\ldots,t_n \in GT \} | ||
| - | \] | + | \end{equation*} |
| Note that if $C$ is true in $I$, then $expand(C)$ is also true in $I$ ($expand(C)$ is a consequence of $C$). | Note that if $C$ is true in $I$, then $expand(C)$ is also true in $I$ ($expand(C)$ is a consequence of $C$). | ||
| We expand entire set: | We expand entire set: | ||
| - | \[ | + | \begin{equation*} |
| expand(S) = \bigcup_{C \in S} expand(C) | expand(S) = \bigcup_{C \in S} expand(C) | ||
| - | \] | + | \end{equation*} |
| Clauses in the expansion have no variables, they are //ground clauses//. | Clauses in the expansion have no variables, they are //ground clauses//. | ||
| Line 31: | Line 31: | ||
| Define propositional model $I_P : V \to \{\it true},{\it false\}$ by | Define propositional model $I_P : V \to \{\it true},{\it false\}$ by | ||
| - | \[ | + | \begin{equation*} |
| I_P(p(C_G)) = e_F(C_G)(I) | I_P(p(C_G)) = e_F(C_G)(I) | ||
| - | \] | + | \end{equation*} |
| Let | Let | ||
| - | \[ | + | \begin{equation*} |
| propExpand(S) = \{ p(C_G) \mid C_G \in expand(S) \} | propExpand(S) = \{ p(C_G) \mid C_G \in expand(S) \} | ||
| - | \] | + | \end{equation*} |
| **Lemma:** If $I$ is a model of $S$, then $I_P$ is a model of $propExpand(S)$. | **Lemma:** If $I$ is a model of $S$, then $I_P$ is a model of $propExpand(S)$. | ||
| Line 66: | Line 66: | ||
| * set $propExpand(S)$ has a propositional model | * set $propExpand(S)$ has a propositional model | ||
| * set $S$ has a model with domain $GT$ | * set $S$ has a model with domain $GT$ | ||
| + | |||
| + | **Example:** Take the first three clauses our example (from [[Normal Forms for First-Order Logic]]): | ||
| + | * $R(x,g(x))$ | ||
| + | * $\neg R(x,y) \lor R(x,f(y,z)))$ | ||
| + | * $P(x) \lor P(f(x,a))$ | ||
| + | Taking as domain $D$ the set of natural numbers, the structure $(D,\alpha)$ is a model of the conjunction of these three clauses, where $\alpha$ is given by: | ||
| + | * $\alpha(a) = 1$ | ||
| + | * $\alpha(c) = 2$ | ||
| + | * $\alpha(g)(x) = (x + 1)$ | ||
| + | * $\alpha(f)(y,z) = y+z$ | ||
| + | * $\alpha(R) = \{(x,y).\ x < y\}$ | ||
| + | * $\alpha(P) = \{x.\ x \mbox{ is even} \}$ | ||
| + | This model induces an Herbrand model $(GT,\alpha_H)$, in which $\alpha_H(P)$ is a set of ground terms and $\alpha_H(R)$ is a relation on ground terms, $\alpha_H(R) \subseteq GT^2$. | ||
| + | |||
| + | To determine, for example, whether $(f(a,a),g(c)) \in \alpha_H(R)$ we check the truth value of the formula | ||
| + | \begin{equation*} | ||
| + | R(f(a,a),g(c))) | ||
| + | \end{equation*} | ||
| + | in the original interpretation $(D,\alpha)$. The truth value of the above formula in $\alpha$ reduces to $1+1 < 2+1$, which is true. Therefore, we define $\alpha_H(R)$ to contain the pair of ground terms $(f(a,a),g(c)))$. On the other hand, $R(f(a,a),c)$ evaluates to false in $(D,\alpha)$, so we define $\alpha_H(R)$ so that it does not contain the pair $(f(a,a),c)$. | ||