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Proof step 12 from proof div-minus-onediff-01 (previous step)

0 < j, j0 < 0 ⊦ j = 0, ((j0 - 1) * j + 1) / j = j0, 1 % j = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 13:
j0 = i0 + 1, 0 < j, j0 < 0 ⊦ j = 0, (i0 * j + 1) / j = j0, 1 % j = 0

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [j0, 1, i] flags:

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: j0, 1, i