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--- sav08:complete_recursive_axiomatizations [2008/04/06 16:51]
vkuncak
+++ sav08:complete_recursive_axiomatizations [2009/05/16 10:47]
vkuncak
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 ====== 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:
   * $Ax$ 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 $A_i$;
@@ Line 9: / Line 8: @@
 **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 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 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.
+    * 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 $F \notin Conseq(Ax)$.
-**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 $F \notin Conseq(Ax)$.
@@ Line 23: / Line 18: @@
 ++++
-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