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Proof step 3 from proof nummult-ls-03 (previous step)

0 < i, j < 0 ⊦ j < i * i0 ↔ (j + 1 - i) / i < i0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
j + 1 = i1 + i, 0 < i, j < 0 ⊦ j < i * i0 ↔ i1 / i < i0

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [j + 1, i, i1] flags:

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: j + 1, i, i1