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Proof step 6 from proof notone (previous step)

i ^ j -1 < i, i ^ j -1 ≤ 0, i ^ j -1 < 1, 2 ≤ i, 1 < i, 0 < j, i ^ j -1 < j, i ^ j -1 + 1 < i, ∀ j0, i. i→n(j0) < i→n(j) → true → 1 < i ∧ 0 < j0 → ¬ i ^ j0 ≤ 1 ⊦

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
j = i0 +1, i ^ i0 < i, i ^ i0 ≤ 0, i ^ i0 < 1, 2 ≤ i, 1 < i, 0 < j, i ^ i0 < j, i ^ i0 + 1 < i, ∀ j0, i. i→n(j0) < i→n(j) → true → 1 < i ∧ 0 < j0 → ¬ i ^ j0 ≤ 1 ⊦

Rule argument

spec: int-basic1 inst: name: elim-pred seq: ⊦ i = j -1 ↔ j = i +1 subst: [j, i] → [j, i0] flags:

Additional information

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