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sav08:compactness_for_first-order_logic [2008/03/20 17:51] vkuncak |
sav08:compactness_for_first-order_logic [2008/03/20 17:58] vkuncak |
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| ====== Compactness for First-Order Logic ====== | ====== Compactness for First-Order Logic ====== | ||
| - | Let $S_0$ be a set of first-order formulas. | + | **Theorem (Compactness for First-Order Logic):** If every finite subset of a set $S_0$ of first-order formulas has a model, then $S_0$ has a model. |
| - | Suppose $S_0$ has no model. | + | **Proof:** |
| + | |||
| + | Let $S_0$ be a set of first-order formulas. We show contrapositive. | ||
| + | |||
| + | Suppose $S_0$ has no model. | ||
| Then $expandProp(clauses(S_0))$ has no model. | Then $expandProp(clauses(S_0))$ has no model. | ||
| - | Some finite subset $S_1 \subseteq expandProp(clauses(S_0))$ of it has no model. | + | Then by [[Compactness Theorem]] for propositional logic, some finite subset $S_1 \subseteq expandProp(clauses(S_0))$ of it has no model. |
| There is then finite subset of clauses $S_2 \subseteq clauses(S_0)$ that generate $S_1$, i.e. such that $S_1 \subseteq expandProp(S_2)$. Therefore, $S_2$ has no model. | There is then finite subset of clauses $S_2 \subseteq clauses(S_0)$ that generate $S_1$, i.e. such that $S_1 \subseteq expandProp(S_2)$. Therefore, $S_2$ has no model. | ||
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| These clauses are generated by a finite subset $S_3 \subseteq S_0$, i.e. $S_2 \subseteq clauses(S_3)$. | These clauses are generated by a finite subset $S_3 \subseteq S_0$, i.e. $S_2 \subseteq clauses(S_3)$. | ||
| - | Therefore $S_3$ has no model. | + | Therefore the finite subset $S_3$ of $S_0$ has no model. |
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| + | **End of Proof.** | ||