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Proof step 11 from proof one-neg (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
1 = i0 * j + j0, i0 + j < ~1, i0 < 0, j0 + j < 0, j < j0, i0 < j0, j < 0, 0 ≤ j0 ⊦ j = ~1, j = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i + j < j ↔ i < 0; (s)%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs% ⊦ i < 0 ∧ j < 0 → i + j < 0; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + (* j) < (* j0) ↔ i < j0 - j; (s)% ⊦ i < ~ j ↔ i + j < 0; (s)% ⊦ 0 - i = ~ i; (s)