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Proof step 2 from proof div-minus-onediff-01 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
i + 1 = i0 + j, i % j = 0, j + i ≤ 0, i < j, i + 1 < j, 0 < j, i < 0 ⊦ j = 1, i0 / j = i / j

Rule argument

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

Additional information

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