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Proof step 2 from proof div-add-nosucc (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

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

Additional information

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