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Both sides previous revision Previous revision Next revision | Previous revision Next revision Both sides next revision | ||
sav08:qe_for_presburger_arithmetic [2008/04/21 21:46] david |
sav08:qe_for_presburger_arithmetic [2008/04/21 21:51] david |
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That's it! | That's it! | ||
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Line 112: | Line 114: | ||
$\exists res, res',i, i'. \neg F$ | $\exists res, res',i, i'. \neg F$ | ||
- | 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 : | ||
$\exists res, i. \neg (res + 2i = 2x \rightarrow res + 2 + 2(i-1) = 2x)$\\ | $\exists res, i. \neg (res + 2i = 2x \rightarrow res + 2 + 2(i-1) = 2x)$\\ | ||
- | $\exists res, i. \neg (res + 2i = 2x \rightarrow res + 2i = 2x)$ | + | $\exists res, i. \neg (res + 2i = 2x \rightarrow res + 2i = 2x)$\\ |
+ | $\exists res, i. \neg true$\\ | ||
+ | $false$ | ||
===== Some Improvements ===== | ===== Some Improvements ===== |