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Proof step 7 from proof div-add1-negpos-neg-02 (previous step)

i % j = 0, i + i0 < 0, i < 0, 0 < j, 0 ≤ i0 ⊦ (i + i0) / j = i / j + (i0 + j - 1) / j, 1 < j ∧ i0 % j = 0, 1 < j ∧ 0 < i0 ∧ i0 % j ≠ 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
i % j = 0, i0 + j = i1 + 1, 0 < j, i < 0, i + i0 < 0, 0 ≤ i0 ⊦ (i + i0) / j = i / j + i1 / j, 1 < j ∧ 0 < i0 ∧ i0 % j ≠ 0, 1 < j ∧ i0 % j = 0

Rule argument

spec: int-abs name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [i0 + j, 1, i1]

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i0 + j, 1, i1