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Proof step 13 from proof div-add1-posneg-pos-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%zero-gels-03: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%sub-add-01: ⊦ i + (j - i0) = i + j - i0; 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%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2