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Proof step 6 from proof ls-pos-02 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
j = i1 + 1, i0 ^ j < i ^ j, i ≤ i0, 0 ≤ i0, 0 ≤ i, 0 ≤ j, ∀ j0, i0, i. i→n(j0) < i→n(j) → i0 ^ j0 < i ^ j0 ∧ i ≤ i0 ∧ 0 ≤ i0 ∧ 0 ≤ i ∧ 0 ≤ j0 → false ⊦ i0 ^ i1 < i ^ i1 ∧ 0 < j

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