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Proof step 8 from proof one-neg (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
1 = i0 * j + j0, i0 < j0, j < j0, j0 < abs(j), i0 < 0, i0 + 1 < abs(j), i0 < abs(j), j < 0, 0 ≤ j0 ⊦ j = 0, j = ~1

Rule argument

Additional information

Used simplifier rules: ⊦ i < abs(i) ↔ i < 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 + 1 < abs(i) ↔ i < 0; (s)% ⊦ i +1 = i + 1; (s)%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% ⊦ 0 < abs(i) ↔ i ≠ 0; (s)% ⊦ ~1 < i ↔ 0 ≤ i; (s)%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%numlef-01: ⊦ (* i) ≤ j → i - 1 < j; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs