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Proof step 10 from proof one-neg (previous step)

j < 0, 1 = i0 * j + j0, i0 < j0, j < j0, 0 ≤ j0, j0 < 0 - j, i0 < 0, i0 + 1 < 0 - j, i0 < 0 - j, j < 0 ⊦ j = 0, j = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
1 = i0 * j + j0, i0 + j < ~1, i0 < 0, j0 + j < 0, 0 ≤ j0, j < j0, i0 < j0, j < 0 ⊦ j = ~1, j = 0

Rule argument

Additional information

Used simplifier rules: ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%zero-01: ⊦ i < 0 ∧ j < 0 → i + j < 0; 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%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs%neg-07: ⊦ i < ~ j ↔ i + j < 0; from specification int-abs%neg-06: ⊦ 0 - i = ~ i; from specification int-abs