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Proof step 20 from proof le-pos (previous step)

2 ≤ i, 1 < i, ~2 < i0, 0 ≤ i0, ~1 ≤ i0, j0 ≤ 0, j0 ≤ i0, ∀ j1, i, j0. i→n(j1) < i→n(i0 +1) → 2 ≤ i ∧ 0 ≤ j1 ∧ 1 < i → (i ^ j1 ≤ i ^ j0 ↔ j1 ≤ j0) ⊦ j0 = 0, i ^ i0 * i ≤ i ^ j0 ↔ i0 + 1 ≤ j0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
j0 ≤ i0, 2 ≤ i, 1 < i, ~2 < i0, 0 ≤ i0, ~1 ≤ i0, j0 < 0, j0 < i0, ∀ j1, i, j0. i→n(j1) < i→n(i0 +1) → 2 ≤ i ∧ 0 ≤ j1 ∧ 1 < i → (i ^ j1 ≤ i ^ j0 ↔ j1 ≤ j0) ⊦ j0 = 0, i ^ i0 * i ≤ 1 / i ^ abs(j0) ↔ i0 + 1 ≤ j0

Rule argument

Additional information

Used simplifier rules: Pot-neg: ⊦ i < 0 → j ^ i = 1 / j ^ abs(i);%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs