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Proof step 23 from proof div-eq-same-01 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 24:
j = i1 + 1, 0 = i1 * i0 + j0, 0 < i0, 0 < j, j0 < j, 0 ≤ j0 ⊦ j = 0, i0 = 0, j = 1

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