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Proof step 29 from proof div-eq-same-01 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 30:
0 = i0 * i + j0, j = i0 + 1, 0 < i, j0 < j, 0 < j, 0 ≤ j0 ⊦ i = 0, j = 1, j = 0

Rule argument

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

Additional information

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