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Proof step 3 from proof zero-gr-add-02 (previous step)

j + 1 < i1 * j + i0, i0 ≤ i1, 0 < i1 * j + i0, 0 ≤ i0, 0 ≤ i1, j < 0, j < i1 * j + i0 ⊦ i1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
0 ≤ i0, 0 < i1 * j + i0, i0 ≤ i1, j + 1 < i1 * j + i0, j < i1 * j + i0, j < 0, 0 < i1, j + 1 < i1 ⊦ i1 = 0

Rule argument

Additional information

Used simplifier rules: succ-01: ⊦ i +1 = i + 1; from specification int-basic2%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs