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Both sides previous revision Previous revision Next revision | Previous revision Next revision Both sides next revision | ||
sav08:simple_qe_for_integer_difference_inequalities [2009/04/21 19:15] vkuncak |
sav08:simple_qe_for_integer_difference_inequalities [2009/04/21 19:17] vkuncak |
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$\forall y \exists x . x \leq y \wedge x \neq y $ | $\forall y \exists x . x \leq y \wedge x \neq y $ | ||
- | ++|\\ | + | ++++|\\ |
$\exists x . x \leq y \wedge (x < y \vee y < x)$\\ | $\exists x . x \leq y \wedge (x < y \vee y < x)$\\ | ||
$\exists x . x \leq y \wedge (x + 1 \leq y \vee y + 1 \leq x)$\\ | $\exists x . x \leq y \wedge (x + 1 \leq y \vee y + 1 \leq x)$\\ | ||
$(\exists x . x \leq y \wedge x + 1 \leq y) \vee (\exists x . x \leq y \wedge y + 1 \leq x)$\\ | $(\exists x . x \leq y \wedge x + 1 \leq y) \vee (\exists x . x \leq y \wedge y + 1 \leq x)$\\ | ||
- | Without lower or upper bounds, formulas is $true$\\ | + | Because there are no lower or upper bounds, these formulas are $true$\\ |
- | $\exists x . x + 0 \leq y \wedge x + 1 \leq y$ is $true$, then the entire formula is $true$ | + | ++++ |
- | ++ | + | |
If we apply this technique to a closed formula in this theory, what formulas do we obtain as a result? | If we apply this technique to a closed formula in this theory, what formulas do we obtain as a result? | ||