Differences

This shows you the differences between two versions of the page.

--- sav08:complete_recursive_axiomatizations [2008/04/06 16:43]
vkuncak
+++ sav08:complete_recursive_axiomatizations [2008/04/06 16:52]
vkuncak
@@ Line 2: / Line 2: @@
 ----
-**Theorem:** Let a set of formulas $Ax$ be a recursive 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 recursive: there exists a an algorithm for checking, given $F$, whether $F \in Ax$
+  * $Ax$ 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 $F \in Conseq(Ax)$.
-----
+**Proof.**++++|
+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 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.
+    * 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 $F \notin Conseq(Ax)$.
-In other words, if a complete theory has a recursive axiomatization, then this theory is decidable.
-(Note: a finite axiomatization is recursive. Typical axiomatizations that use "axiom schemas" are also recursive.)
+**End of Proof.**
+++++
-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.
+In other words, if a complete theory has a recursively enumerable axiomatization, then this theory is decidable.
-**Proof.**
+(Note: a finite axiomatization is recursively enumerable. Typical axiomatizations that use "axiom schemas" are also recursively enumerable.)
-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.
+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.
-**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.  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.
-  * 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 $F \notin Conseq(Ax)$.
-**End of Proof.**
 Example: the theory of integers with multiplication and quantifiers is undecidable