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sav08:complete_recursive_axiomatizations [2008/04/06 16:52] vkuncak |
sav08:complete_recursive_axiomatizations [2009/05/16 10:47] vkuncak |
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| ====== Complete Recursive Axiomatizations ====== | ====== Complete Recursive Axiomatizations ====== | ||
| - | ---- | ||
| **Theorem:** Let a set of first-order sentences $Ax$ be a recursively enumerable axiomatization for a [[First-Order Theories|complete and consistent theory]], that is: | **Theorem:** Let a set of first-order sentences $Ax$ be a recursively enumerable axiomatization for a [[First-Order Theories|complete and consistent theory]], that is: | ||
| * $Ax$ is recursively enumerable: there exists an enumerateion $A_1,A_2,\ldots$ of the set $Ax$ and there exists an algorithm that given $i$ computes $A_i$; | * $Ax$ is recursively enumerable: there exists an enumerateion $A_1,A_2,\ldots$ of the set $Ax$ and there exists an algorithm that given $i$ computes $A_i$; | ||
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| **Proof.**++++| | **Proof.**++++| | ||
| Suppose $Ax$ is a complete recursive axiomatization. There are two cases, depending on whether $Ax$ is consistent. | Suppose $Ax$ is a complete recursive axiomatization. There are two cases, depending on whether $Ax$ is consistent. | ||
| - | * **Case 1):** The set $Ax$ is inconsistent, that is, there are not models for $Ax$. Then $Conseq(Ax)$ is the set of all first-order sentences and there is a trivial algorithm for checking whether $F \in Conseq(Ax)$: always return true. | + | * **Case 1):** The set $Ax$ is inconsistent, that is, there are no models for $Ax$. Then $Conseq(Ax)$ is the set of all first-order sentences and there is a trivial algorithm for checking whether $F \in Conseq(Ax)$: always return true. |
| * **Case 2):** The set $Ax$ is consistent. Given $F$, to check whether $F \in Conseq(Ax)$ we run in parallel two complete theorem provers (which exist by [[lecture10|Herbrand theorem]]), the first one trying to prove the formula $F$, the second one trying to prove the formula $\lnot F$; the procedure terminates if any of these theorem provers succeed (the theorem provers simultaneously searches for longer and longer proofs from a larger and larger finite subsets of $Ax$). We show that this is an algorithm that decides $F \in Conseq(Ax)$. | * **Case 2):** The set $Ax$ is consistent. Given $F$, to check whether $F \in Conseq(Ax)$ we run in parallel two complete theorem provers (which exist by [[lecture10|Herbrand theorem]]), the first one trying to prove the formula $F$, the second one trying to prove the formula $\lnot F$; the procedure terminates if any of these theorem provers succeed (the theorem provers simultaneously searches for longer and longer proofs from a larger and larger finite subsets of $Ax$). We show that this is an algorithm that decides $F \in Conseq(Ax)$. | ||
| * Because either $F \in Conseq(Ax)$ or $(\lnot F) \in Conseq(Ax)$, and theorem provers are complete, one of these theorem provers will eventually halt. The procedure is therefore an algorithm. | * Because either $F \in Conseq(Ax)$ or $(\lnot F) \in Conseq(Ax)$, and theorem provers are complete, one of these theorem provers will eventually halt. The procedure is therefore an algorithm. | ||
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| ++++ | ++++ | ||
| - | In other words, if a complete theory has a recursively enumerable axiomatization, then this theory is decidable. | + | In short: if a complete theory has a recursively enumerable axiomatization, then this theory is decidable. |
| - | + | ||
| - | (Note: a finite axiomatization is recursively enumerable. Typical axiomatizations that use "axiom schemas" are also recursively enumerable.) | + | |
| Conversely: if a theory is undecidable (there is no algorithm for deciding whether a sentence is true or false), then the theory does not have a recursive axiomatization. | Conversely: if a theory is undecidable (there is no algorithm for deciding whether a sentence is true or false), then the theory does not have a recursive axiomatization. | ||
| + | Note: | ||
| + | * a finite axiomatization is recursively enumerable | ||
| + | * typical axiomatizations that use "axiom schemas" are also recursively enumerable | ||
| + | * if $Ax$ is an axiomatization for for some interpretation $I$, that is $Conseq(Ax) = \{ F \mid e_F(F)(I) \}$, then $Ax$ is an axiomatization of a complete theory | ||
| + | |||
| + | **Corrollary:** Let $I$ be an interpretation. Then exactly one of the following is true: | ||
| + | * there exists an algorithm for checking $e_F(F)(I)$ | ||
| + | * there is no enumerable set of axioms $Ax$ such that $Conseq(Ax) = \{ F \mid e_F(I) \}$. | ||
| Example: the theory of integers with multiplication and quantifiers is undecidable | Example: the theory of integers with multiplication and quantifiers is undecidable | ||