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Proof step 9 from proof mod-nls-04 (previous step)

¬ i0 + j ≠ 1, i0 < 0, abs(i % j) < j, i % j < i0, i0 + j ≤ 1, 0 < j ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: mod-less-01: ⊦ 0 < j → i % j < j;%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%add-less: ⊦ 0 < i → i1 < i1 + i; from specification int-basic2%ls-01: ⊦ i + j < i + i0 ↔ j < i0; from specification int-basic2%lst-04: ⊦ i + j < i0 ∧ j0 < j → i + j0 < i0; from specification int-basic2%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%neg-13: ⊦ ~ i < j ↔ 0 < j + i; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%forward-le: ⊦ i ≤ j ∧ 0 < i → 0 < j; (forward) from specification int-abs