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sav08:complete_recursive_axiomatizations [2008/04/06 16:49] vkuncak |
sav08:complete_recursive_axiomatizations [2008/04/06 16:51] vkuncak |
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| - | **Theorem:** Let a set of formulas (axioms) $Ax$ 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 $A_i$; | ||
| * $Conseq(Ax)$ is complete: for each FOL sentence $F$, either $F \in Conseq(Ax)$ or $(\lnot F) \in Conseq(Ax)$; | * $Conseq(Ax)$ is complete: for each FOL sentence $F$, either $F \in Conseq(Ax)$ or $(\lnot F) \in Conseq(Ax)$; | ||
| Then there exists an algorithm for checking, given $F$, whether $F \in Conseq(Ax)$. | Then there exists an algorithm for checking, given $F$, whether $F \in Conseq(Ax)$. | ||
| - | ---- | + | **Proof.**++++| |
| - | + | ||
| - | In other words, 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. | + | |
| - | + | ||
| - | **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 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 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 $F \in Conseq(Ax)$. | ||
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| * If $F$ is proved, then by soundness of theorem prover, $F \in Conseq(Ax)$. | * If $F$ is proved, then by soundness of theorem prover, $F \in Conseq(Ax)$. | ||
| * If $(\lnot F)$ is proved, then by soundness of theorem prover $(\lnot F) \in Conseq(Ax)$. Because $Ax$ is consistent, there is a model $I$ for $Ax$. Then $(\lnot F)$ is true in $I$, so $F$ is false in $I$. Because there is a model where $F$ does not hold, we have $F \notin Conseq(Ax)$. | * If $(\lnot F)$ is proved, then by soundness of theorem prover $(\lnot F) \in Conseq(Ax)$. Because $Ax$ is consistent, there is a model $I$ for $Ax$. Then $(\lnot F)$ is true in $I$, so $F$ is false in $I$. Because there is a model where $F$ does not hold, we have $F \notin Conseq(Ax)$. | ||
| + | |||
| **End of Proof.** | **End of Proof.** | ||
| + | ++++ | ||
| + | |||
| + | In other words, 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. | ||
| + | |||
| Example: the theory of integers with multiplication and quantifiers is undecidable | Example: the theory of integers with multiplication and quantifiers is undecidable | ||