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Proof step 16 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 + 1) * j + (j1 - j0 - j) ∧ 0 ≤ 0 ∧ 0 ≤ j1 - j0 - j ∧ j1 - j0 - j < j, j0 = 0, i0 + i1 = i2 + 1, i2 * j + j1 = (i0 + i1) * j + j0 ∧ j1 < j0 + j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: add-mult-two: ⊦ i + i = 2 * i; from specification int-mult%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%eq-succ: ⊦ i * (i0 + 1) + j = i + j0 ↔ i * i0 + j = j0; from specification int-mult%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%sub-sub: ⊦ 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