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Proof step 8 from proof le-pos (previous step)

0 < j, 2 ≤ i, 0 ≤ j, 1 < i, ∀ j1, i, j0. i→n(j1) < i→n(j) → 2 ≤ i ∧ 0 ≤ j1 ∧ 1 < i → (i ^ j1 ≤ i ^ j0 ↔ j1 ≤ j0) ⊦ i ^ j -1 * i ≤ i ^ j0 ↔ j ≤ j0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
j = i0 +1, 0 < j, 2 ≤ i, 0 ≤ j, 1 < i, ∀ j1, i, j0. i→n(j1) < i→n(j) → 2 ≤ i ∧ 0 ≤ j1 ∧ 1 < i → (i ^ j1 ≤ i ^ j0 ↔ j1 ≤ j0) ⊦ i ^ i0 * i ≤ i ^ j0 ↔ j ≤ j0

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