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Proof step 9 from proof le-pos-01 (previous step)

0 < j, 0 ≤ i0, 0 ≤ i, i ^ j ≤ i0 ^ j, i0 < i, 0 ≤ j, 0 < i, ∀ j0, i0, i. i→n(j0) < i→n(j) → 0 < j0 ∧ 0 ≤ i0 ∧ 0 ≤ i ∧ i ^ j0 ≤ i0 ^ j0 ∧ i0 < i ∧ 0 ≤ j0 ∧ 0 < i → false ⊦ 1 < j ∧ i ^ (j - 1) ≤ i0 ^ (j - 1)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
j = i1 + 1, 0 < j, 0 ≤ i0, 0 ≤ i, i ^ j ≤ i0 ^ j, i0 < i, 0 ≤ j, 0 < i, ∀ j0, i0, i. i→n(j0) < i→n(j) → 0 < j0 ∧ 0 ≤ i0 ∧ 0 ≤ i ∧ i ^ j0 ≤ i0 ^ j0 ∧ i0 < i ∧ 0 ≤ j0 ∧ 0 < i → false ⊦ 1 < j ∧ i ^ i1 ≤ i0 ^ i1

Rule argument

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

Additional information

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