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--- sav07_lecture_23 [2007/07/02 17:14]
simon.blanchoud
+++ sav07_lecture_23 [2008/05/21 23:14]
vkuncak
@@ Line 111: / Line 111: @@
 Modularity: [[http://doi.acm.org/10.1145/316686.316703|Componential set-based analysis]]
@@ Line 123: / Line 124: @@
   * reduction to term algebras ({{http://theory.stanford.edu/~aiken/publications/papers/popl02.pdf|First-order theory of structural subtyping}})
-Deciding monadic class (({{bachmairetal93setconstraintsmonadicclass.ps|Set constraints are the monadic class}}))
+==== Term Algebras ====
-  * reduction to boolean algebras (MSOL fragment, has QE)
-  * small model property
+  * $L = \lbrace a, f, \ldots \rbrace$
-  * resolution variants
+  * $H$ : set of all ground terms
+  * First-order theory : $\forall x \exist y . f(f(a,a), x) = f(y, a) \wedge y \neq a$
+This is First-Order Logic (FOL) with variables ! We can relate this to set constraints
+==== Monadic Class of FOL ====
+ MFOL(({{bachmairetal93setconstraintsmonadicclass.ps|Set constraints are the monadic class}}))  has : only unary predicates, no function symbol, quatifiers and logic.
+Translating this to our sets language :
+  F := S = S
+     | S ⊆ S
+     | card(S) ≥ k
+     | card(S) = k
+     | F ∧ F
+     | ¬ F
+     | ∃v.F
+  S := v
+     | S ∪ S
+     | S ∩ S
+     | S'
+     | ∅
+     | 1
+ Where :
+  * $card(S) \geq k$ is syntactic sugar to represent a unary operator, as $k$ is not a variable, that compare the cardinality of $S$ with $k$
+  * $k \geq 0$ is an integer
+  * $S'$ is the complement set of $S$
+  * $1$ represents the whole universe
+ In order to agree with our definitions of set constraints, we need to eliminate :
+  * $S_1 = S_2 \rightarrow S_1 \subseteq S_2 \wedge S_2 \subseteq S_1 $
+  * $S_1 \subseteq S_2 \rightarrow card(S_1 \cup S_2') = 0$
+We then have to be able to represent the following formulas :
+  * $\neg(card(s) \geq k) \rightarrow \bigvee_{i=0}^{k-1} card(S) = i$
+  * $\neg(card(s) = k) \rightarrow card(S) \geq k+1 \vee \bigvee_{i=0}^{k-1} card(S) = i$
+We can then perform quantifier elimination. Note that this is not possible in case we do not have the $card$ operator : \[\exists v.card(S_1 \cap v) = k_1 \wedge card(S_1 \cap v') = k_2 \Rightarrow card(S_1) = k_1 + k_2\]
+After elimination of the quantifiers, deciding the result of the formula is trivial.