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Proof step 17 from proof div-eq-same-01 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

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

Additional information

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