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msol_over_strings [2007/05/27 12:43]
vaibhav.rajan
msol_over_strings [2007/05/27 13:34]
vaibhav.rajan
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   [\![\exists v. F]\!]e &=& \exists S. S\ \mbox{is finite}\ \land\ S \subseteq N_0 \land [\![F]\!](e[v \mapsto S])   [\![\exists v. F]\!]e &=& \exists S. S\ \mbox{is finite}\ \land\ S \subseteq N_0 \land [\![F]\!](e[v \mapsto S])
 \end{eqnarray*} \end{eqnarray*}
 +
 +
 +
  
  
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   * Intersection:​ $(A = B \cap C) = (A \subseteq B \land A \subseteq C) \land (\forall A_1. A_1 \subseteq B \land A_1 \subseteq C \rightarrow A_1 \subseteq A)$   * Intersection:​ $(A = B \cap C) = (A \subseteq B \land A \subseteq C) \land (\forall A_1. A_1 \subseteq B \land A_1 \subseteq C \rightarrow A_1 \subseteq A)$
   * Union: $(A = B \cup C) = (B \subseteq A \land C \subseteq A) \land (\forall A_1. B \subseteq A_1 \land C \subseteq A_1 \rightarrow A \subseteq A_1)$   * Union: $(A = B \cup C) = (B \subseteq A \land C \subseteq A) \land (\forall A_1. B \subseteq A_1 \land C \subseteq A_1 \rightarrow A \subseteq A_1)$
-  * Set difference: $(A = B \setminus C) = (A \cup C = B \land A \cap C = \emptyset)$+  * Set difference: $(A = B \setminus C) = (A \cup (B \cap C= B \land A \cap C = \emptyset)$
     (or just use element-wise definitions with singletons)     (or just use element-wise definitions with singletons)
   * If $k$ is a fixed constant, properties $\mbox{card}(A) \geq k$, $\mbox{card}(A)\leq k$, $\mbox{card}(A)=k$   * If $k$ is a fixed constant, properties $\mbox{card}(A) \geq k$, $\mbox{card}(A)\leq k$, $\mbox{card}(A)=k$
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 \end{equation*} \end{equation*}
  
 +This does not give the smallest set containing both $u$ and $v$. The reflexive transitive closure, T is:
 +\begin{equation*}
 +(F \subseteq T )\land (\forall x. x \in S \rightarrow (x,x) \in T ) \land ((\exists k.(u,k) \in T \land (k,v) \in T) \rightarrow ((u,v) \in T))
 +\end{equation*}
 +The underlying smallest set S containing $u$ and $v$ is given by:
 +\begin{equation*}
 +S = \{x | \exists k. ((k, x) \in T \lor (x,k) \in T) \}
 +\end{equation*} ​
 **Using transitive closure and successors:​** **Using transitive closure and successors:​**
   * Constant zero: $(x=0) = One(x) \land \lnot (\exists y. One(y) \land s(y,x))$   * Constant zero: $(x=0) = One(x) \land \lnot (\exists y. One(y) \land s(y,x))$
   * Addition by constant: $(x = y + c) = (\exists y_1,​\ldots,​y_{c-1}. s(y,y_1) \land s(y_1,y_2) \land \ldots \land s(y_{c-1},​x))$   * Addition by constant: $(x = y + c) = (\exists y_1,​\ldots,​y_{c-1}. s(y,y_1) \land s(y_1,y_2) \land \ldots \land s(y_{c-1},​x))$
   * Ordering on positions in the string: $(u \leq v) = ((u,v) \in \{(x,​y)|s(x,​y))\}^*$   * Ordering on positions in the string: $(u \leq v) = ((u,v) \in \{(x,​y)|s(x,​y))\}^*$
-  * Reachability in $k$-increments,​ that is, $\exists k \geq 0. y=x + c\cdot k$: $\mbox{Reach}_c(u,​v) = ((u,v) \in \{(x,y)\mid y=x+c\})$+  * Reachability in $k$-increments,​ that is, $\exists k \geq 0. y=x + c\cdot k$: $\mbox{Reach}_c(u,​v) = ((u,v) \in \{(x,y)\mid y=x+c\}^*)$
   * Congruence modulo $c$: $(x \equiv y)(\mbox{mod}\ c) = \mbox{Reach}_c(x,​y) \lor \mbox{Reach}_c(y,​x)$   * Congruence modulo $c$: $(x \equiv y)(\mbox{mod}\ c) = \mbox{Reach}_c(x,​y) \lor \mbox{Reach}_c(y,​x)$