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sav08:qe_for_presburger_arithmetic [2009/04/23 09:57] vkuncak |
sav08:qe_for_presburger_arithmetic [2012/03/09 10:44] vkuncak |
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We consider elimination of a quantifier from a conjunction of literals (because [[QE from Conjunction of Literals Suffices]]). | We consider elimination of a quantifier from a conjunction of literals (because [[QE from Conjunction of Literals Suffices]]). | ||
+ | Running example: | ||
+ | \[ | ||
+ | \exists y. 3 y - 2 x + 1 > - x \land 2y - 6 < z \land 4 \mid 5y + 1 | ||
+ | \] | ||
===== Normalizing Conjunctions of Literals ===== | ===== Normalizing Conjunctions of Literals ===== | ||
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We obtain a disjunction of conjunctions of literals of the form $0 < t$ and $K \mid t$ where $t$ are of the form $K_0 + \sum_{i=1}^n K_i \cdot x_i$ | We obtain a disjunction of conjunctions of literals of the form $0 < t$ and $K \mid t$ where $t$ are of the form $K_0 + \sum_{i=1}^n K_i \cdot x_i$ | ||
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- | |||
===== Exposing the Variable to Eliminate ===== | ===== Exposing the Variable to Eliminate ===== | ||
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We first drop all constraints except divisibility, obtaining $F_2(x)$ | We first drop all constraints except divisibility, obtaining $F_2(x)$ | ||
\[ | \[ | ||
- | \bigwedge_{i=1}^D K_i \mid (x_i + t_i) | + | \bigwedge_{i=1}^D K_i \mid (x + t_i) |
\] | \] | ||
and then eliminate quantifier as | and then eliminate quantifier as |