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sav08:qe_for_presburger_arithmetic [2008/04/21 21:43] david |
sav08:qe_for_presburger_arithmetic [2009/04/21 19:38] 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]]). | ||
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| ===== Normalizing Conjunctions of Literals ===== | ===== Normalizing Conjunctions of Literals ===== | ||
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| $t_1 < t_2$ becomes ++| $0 < t_2 - t_1$++ | $t_1 < t_2$ becomes ++| $0 < t_2 - t_1$++ | ||
| - | We obtain a disjunction of conjunctions of literals of the form $0 < t$ and $K \mid t$ where $t$ are | + | We obtain a disjunction of conjunctions of literals of the form $0 < t$ and $K \mid t$ where $t$ are of the form $\sum_{i=1}^n K_i \cdot x_i$ |
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| That's it! | That's it! | ||
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| $\exists res, res',i, i'. \neg F$ | $\exists res, res',i, i'. \neg F$ | ||
| - | We can eliminate quantifiers with equalities: $i'= i-1$ | + | We can eliminate quantifiers with equalities: $i'= i-1$ and $res' = res + 2$ |
| - | $res' + 2i'$ becomes $res' + 2(i-1)$, and $\exists i'$ can be removed | + | $res' + 2i'$ becomes $res + 2 + 2(i-1)$, and $\exists i', res' $ can be removed |
| Finally : | Finally : | ||
| - | $\exists res, i. \neg (res + 2i = 2x \rightarrow res + 2 + 2(i-1) = 2x)$ | + | $\exists res, i. \neg (res + 2i = 2x \rightarrow res + 2 + 2(i-1) = 2x)$\\ |
| + | $\exists res, i. \neg (res + 2i = 2x \rightarrow res + 2i = 2x)$\\ | ||
| + | $\exists res, i. \neg true$\\ | ||
| + | $false$ | ||
| ===== Some Improvements ===== | ===== Some Improvements ===== | ||