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Proof step 14 from proof mod-add1-negpos-neg (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
j * (i2 + i1) + j0 = i0 * j + j1, abs(j1) < j, 1 < j, i0 * j + j1 < 0, i1 < 0, j0 < j, 0 < j0, 0 < j, i0 < j0, i0 < j, j1 < j0, j1 < j, j1 + j < j0, i0 ≤ 0, j1 ≤ 0, 0 ≤ i2 * j + j0, 0 ≤ i2, j1 ≤ j0 ⊦ j0 = j1 + j, j0 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i < j → ¬ j < i; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ i - j < i0 ↔ i < j + i0; (s)% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; (s)% ⊦ i - j = i0 ↔ i = i0 + j; (s)% ⊦ i + 0 = i; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + (j - i0) = i + j - i0; (s)