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msol_over_strings [2007/05/10 18:15]
vkuncak
msol_over_strings [2007/05/27 12:52]
vaibhav.rajan
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 ====== MSOL over Strings ====== ====== MSOL over Strings ======
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   [\![F_1 \lor F_2]\!]e &=& [\![F_1]\!]e\ \lor\ [\![F_2]\!]e \\ \   [\![F_1 \lor F_2]\!]e &=& [\![F_1]\!]e\ \lor\ [\![F_2]\!]e \\ \
   [\![\lnot F]\!]e &=& \lnot ([\![F]\!]e) \\ \   [\![\lnot F]\!]e &=& \lnot ([\![F]\!]e) \\ \
-  [\![\exists v. F]\!]e &=& \exists S. S\ \mbox{is finite}\ \land\ S \subseteq N_0[\![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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 +
  
 ===== What can we express in MSOL over strings ===== ===== What can we express in MSOL over strings =====
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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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 **Transitive closure of a relation.** If $F(x,y)$ is a formula on singletons, we define reflexive transitive closure as follows. ​ Define shorthand **Transitive closure of a relation.** If $F(x,y)$ is a formula on singletons, we define reflexive transitive closure as follows. ​ Define shorthand
 \begin{equation*} \begin{equation*}
-  \mbox{Closed}(S,​R) = (\forall x,y. One(x) \land One(y) \land x \in S \land F(x,y) \rightarrow y \in S)+  \mbox{Closed}(S,​F) = (\forall x,y. One(x) \land One(y) \land x \in S \land F(x,y) \rightarrow y \in S)
 \end{equation*} \end{equation*}
 Then $(u,v) \in \{(x,y) \mid F(x,y)\}^*$ is defined by Then $(u,v) \in \{(x,y) \mid F(x,y)\}^*$ is defined by
 \begin{equation*} \begin{equation*}
-  \forall S. u \in S \land \mbox{Closed}(S,​R) \rightarrow v \in S+  \forall S. u \in S \land \mbox{Closed}(S,​F) \rightarrow v \in S
 \end{equation*} \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: $(x = y + c) = (\exists y_1,​\ldots,​y_{c-1}. s(y,y_1) \land s(y_1,y_2) \land \ldots \land s(y_{k-1},x))$+  * 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, $\exists k \geq 0. y=x + c\cdot k$: $\mbox{Reach}_c(u,​v) = ((u,v) \in \{(x,y)\mid y=x+c\})$