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Proof step 6 from proof mod-pred-neg-01 (previous step)

i0 * j + j0 = i1 + 1, i0 * j + j0 < j, i0 * j + j0 < 0, i0 < j, i0 ≤ 0, j0 < j, j0 ≤ 0, (i0 * j + j0) + 1 < j, 0 < j, abs(j0) < j ⊦ j0 = i1 % j + 1, i1 % j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
i0 * j + j0 = i1 + 1, i1 < j, abs(j0) < j, 0 < j, j0 ≤ 0, j0 < j, i0 ≤ 0, i0 < j, 2 + i1 < j, i1 < ~1, i1 < 0 ⊦ i1 % j = 0, j0 = i1 % j + 1

Rule argument

Additional information

Used simplifier rules: numlsf: ⊦ i < (* j) ∧ j < 0 → i < 0; (forward) from specification int-basic1%lst-03: ⊦ i + j < i0 ∧ ¬ j < j0 → i + j0 < i0; from specification int-basic2%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs%lst-04: ⊦ i + j < i0 ∧ j0 < j → i + j0 < i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%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