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Proof step 3 from proof div-minus-onediff-01 (previous step)

i % j = 0, 0 < j, i < 0 ⊦ (i + 1 - j) / j = i / j

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
i % j = 0, i + 1 = i0 + j, i < 0, 0 < j ⊦ i0 / j = i / j

Rule argument

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

Additional information

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