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--- sav08:definition_of_set_constraints [2008/05/22 11:41]
vkuncak
+++ sav08:definition_of_set_constraints [2008/05/22 11:55]
vkuncak
@@ Line 104: / Line 104: @@
 Is every set expression in set constraints monotonic? ++|Set difference is not monotonic in second argument.++
-===== Conditional Constraints =====
-An important property we would like to have in this semantic is a very intuitive one : \[ [[f^{-1}(f(S_1, S_2))]] = [[S_1]] \]
+===== Conditional Constraints =====
-This property is in fact not conserved as we can easily see using a very simple counter-example :
-\[ [[f^{-1}(f(S_1, \emptyset))]] = [[f^{-1}(\lbrace f(t_1, t_2) | t_1 \in [[S_1]] \wedge t_2 \in \emptyset \rbrace)]] = [[f^{-1}(\emptyset)]] = \emptyset \]
-The correct interpretation of this property is :
+Let us simplify this expression:
-\begin{displaymath} [[f^{-1}(f(S_1, S_2))]] = \left\{ \begin{array}{ll}
+\[
- \emptyset & if [[S_2]] = \emptyset &
+    f^{-1}(f(S,T)) =
-[[S_1]] & otherwise \\
+\]
-\end{array} \right \end{displaymath}
+++|
+\[
+   = f^{-1}(\{ f(s,t) \mid s \in S, t \in T \})
+\]
+++
+++|
+\[
+   = \{ s_1 \in S \mid \exists t_1 \in T \}
+\]
+++
+++|
+\[
+   = \left\{ \begin{array}{rl}
+       S, \mbox{ if } T \neq \emptyset \\
+       \emptyset, \mbox{ if } T = \emptyset
+              \end{array}
+     \right.
+\]
+++
+Consider conditional constraints of the form
+\[
+    T \neq \emptyset \rightarrow S_1 \subseteq S_2
+\]
+We can transform it into
+\[
+    f(S_1,T) \subseteq f(S_2,T)
+\]
+for some binary function symbol $f$.   Thus, we can introduce such conditional constraints without changing expressive power or decidability.