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msol_over_strings [2007/05/06 19:03] vkuncak created |
msol_over_strings [2007/05/27 12:43] vaibhav.rajan |
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| + | ====== MSOL over Strings ====== | ||
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| [\![v_1 \subseteq v_2]\!]e &=& (e(v_1) \subseteq e(v_2)) \\ \ | [\![v_1 \subseteq v_2]\!]e &=& (e(v_1) \subseteq e(v_2)) \\ \ | ||
| [\![s(v_1,v_2)]\!]e &=& (\exists k \in N_0. e(v_1) = \{k\} \land e(v_2)=\{k+1\}) \\ \ | [\![s(v_1,v_2)]\!]e &=& (\exists k \in N_0. e(v_1) = \{k\} \land e(v_2)=\{k+1\}) \\ \ | ||
| - | [\![F_1 \lor F_2]\!]e &=& [\![F_1]\!]e\ \land\ [\![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]\!] &=& \exists 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 ===== | ||
| **Set operations**. The ideas is that quantification over sets with $\subseteq$ gives us the full Boolean algebra of sets. | **Set operations**. The ideas is that quantification over sets with $\subseteq$ gives us the full Boolean algebra of sets. | ||
| - | * Two sets are equal: $(S_1 = S_2) = (S_1 \subeteq S_2) \land (S_2 \subseteq S_1)$ | + | * Two sets are equal: $(S_1 = S_2) = (S_1 \subseteq S_2) \land (S_2 \subseteq S_1)$ |
| * Strict subset: $(S_1 \subset S_2) = (S_1 \subseteq S_2) \land \lnot (S_2 \subseteq S_1)$ | * Strict subset: $(S_1 \subset S_2) = (S_1 \subseteq S_2) \land \lnot (S_2 \subseteq S_1)$ | ||
| * Set is empty: $(S=\emptyset) = \forall S_1. S \subseteq S_1$ | * Set is empty: $(S=\emptyset) = \forall S_1. S \subseteq S_1$ | ||
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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, $\exists k \geq 0. y=x + c\cdot k$: $\mbox{Reach}_c(u,v) = ((u,v) \in \{(x,y)\mid y=x+c\})$ | ||