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Proof step 6 from proof ls-pos (previous step)

i * j1 < i * (j0 - 1) ↔ j1 + 1 < j0, 0 < i, ∀ i, j0. 0 < i → (i * j1 < i * j0 ↔ j1 < j0) ⊦ i * j1 + i < i * j0 ↔ j1 + 1 < j0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
j0 = i0 + 1, i * j1 < i * i0 ↔ j1 + 1 < j0, 0 < i, ∀ i, j0. 0 < i → (i * j1 < i * j0 ↔ j1 < j0) ⊦ i * j1 + i < i * j0 ↔ j1 + 1 < j0

Rule argument

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

Additional information

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