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sav08:deciding_quantifier-free_fol_over_ground_terms [2008/04/23 15:10] vkuncak |
sav08:deciding_quantifier-free_fol_over_ground_terms [2008/05/19 14:04] vkuncak |
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| Can we treat variables as constants in this case? | Can we treat variables as constants in this case? | ||
| - | Is the following formula satisfiable | + | Is the following formula satisfiable ? |
| \[ | \[ | ||
| a=f(a) \land a \neq b | a=f(a) \land a \neq b | ||
| - | \] | + | \] ++|No. When interpreting variables as constants, two syntactically different terms are different ($a \neq f(a)$).++ |
| - | is the following | + | |
| + | |||
| + | And the following ? | ||
| \[ | \[ | ||
| (f(a)=b \lor (f(f(f(a))) = a \land f(a) \neq f(b))) \land f(f(f(f(a)) \neq b \land b=f(b) | (f(a)=b \lor (f(f(f(a))) = a \land f(a) \neq f(b))) \land f(f(f(f(a)) \neq b \land b=f(b) | ||
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| **Theorem:** Let $E$ be a set of equations and disequations where disequations are only between variables. Let $E^+ = \{ (t_1 \doteq t_2) \mid (t_1 = t_2) \in E \}$ the unification problem for equations in $E$. Then | **Theorem:** Let $E$ be a set of equations and disequations where disequations are only between variables. Let $E^+ = \{ (t_1 \doteq t_2) \mid (t_1 = t_2) \in E \}$ the unification problem for equations in $E$. Then | ||
| * if $E^+$ has no unifier, the set $E$ is unsatisfiable over ground terms; | * if $E^+$ has no unifier, the set $E$ is unsatisfiable over ground terms; | ||
| - | * if $E^+$ has a unifier, if the language contains at least one function symbol of arity at least one (i.e. the set of ground terms is infinite) then $E$ is satisfiable. | + | * if $E^+$ has a most general unifier $\sigma$ such that for all $(x \neq y) \in E$ we have $\sigma(x) \neq \sigma(y)$, then if the language contains at least one function symbol of arity at least one (i.e. the set of ground terms is infinite) then $E$ is satisfiable. |
| Note: we need assumption that there are infinitely many terms, consider for example ${\cal L} = \{ a,b\}$ and formula $x \neq y \land x \neq z \land y \neq z$. | Note: we need assumption that there are infinitely many terms, consider for example ${\cal L} = \{ a,b\}$ and formula $x \neq y \land x \neq z \land y \neq z$. | ||
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| * if we have predicates: this is satisfiability in Herbrand models, equivalent to satisfiability in FOL without equality, so it is undecidable | * if we have predicates: this is satisfiability in Herbrand models, equivalent to satisfiability in FOL without equality, so it is undecidable | ||
| + | ===== References ===== | ||
| + | |||
| + | * [[http://lara.epfl.ch/~kuncak/papers/KuncakRinard03TheoryStructuralSubtyping.html|On the Theory of Structural Subtyping]], see | ||