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Proof step 10 from proof div-add1-posneg-pos-01 (previous step)

i0 * j + i1 * j + j0 = i2 * j + j1 ∧ 0 ≤ j1 ∧ j1 < j ∧ 0 ≤ i2, 0 ≤ i0, j0 + 1 < i0 * j, (i1 * j + j0) + 1 < i0 * j, i1 * j + j0 < 0, i1 < j + j0, i1 < j, i1 ≤ 0, (i1 * j + j0) + 1 < j, 0 < j + j0, j0 + 1 < j, j0 < j, j0 < 0, i1 * j + j0 < j, i1 < i0 * j, 1 < j, 0 < i0, (i1 * j + j0) + 1 < i0, 0 < j, 0 ≤ i0 * j + i1 * j + j0, i1 < i0, j0 + 1 < i0 ⊦ i0 + i1 = i2 + 1, j0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
i0 * j + i1 * j + j0 = i2 * j + j1, j0 < i2 * j + j1, i1 ≤ i2 * j + j1, i1 * j + j0 < i2 * j + j1, 0 ≤ i2, 0 ≤ j1, j0 + 1 < i0, i1 < i0, 0 < j, (i1 * j + j0) + 1 < i0, 0 < i0, 1 < j, i1 < i0 * j, i1 * j + j0 < j, j0 < 0, j0 < j, j0 + 1 < j, 0 < j + j0, (i1 * j + j0) + 1 < j, i1 ≤ 0, i1 < j, i1 < j + j0, i1 * j + j0 < 0, (i1 * j + j0) + 1 < i0 * j, j0 + 1 < i0 * j, 0 ≤ i0, j1 < j ⊦ j0 = 0, i0 + i1 = i2 + 1

Rule argument

Additional information

Used simplifier rules: zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%zero-ge-add: ⊦ 0 ≤ j1 ∧ j1 < j → (0 ≤ i0 * j + j1 ↔ 0 ≤ i0); from specification int-mult%forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs