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Proof step 14 from proof le-pos-01 (previous step)

0 < i, i0 < i, 0 ≤ i1, i ^ i1 * i ≤ i0 ^ i1 * i0, 0 ≤ i0, 0 ≤ i, ∀ j0, i, i0. i→n(j0) < i→n(i1 + 1) → 0 < i ∧ i0 < i ∧ 0 < j0 ∧ i ^ j0 ≤ i0 ^ j0 ∧ 0 ≤ j0 ∧ 0 ≤ i0 ∧ 0 ≤ i → false ⊦ 0 < i1 ∧ i ^ i1 ≤ i0 ^ i1, i0 ^ i1 ≤ i ^ i1 ∧ i0 < i ∧ i ^ i1 ≠ 0 ∧ 0 ≤ i0 ^ i1 ∧ 0 ≤ i0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
i0 < i, 0 < i, 0 ≤ i0 ^ i1, 0 ≤ i, 0 ≤ i0, i ^ i1 * i ≤ i0 ^ i1 * i0, 0 ≤ i1, ∀ j0, i, i0. i→n(j0) < i→n(i1 + 1) → 0 < i ∧ i0 < i ∧ 0 < j0 ∧ i ^ j0 ≤ i0 ^ j0 ∧ 0 ≤ j0 ∧ 0 ≤ i0 ∧ 0 ≤ i → false ⊦ i ^ i1 ≠ 0 ∧ i0 ^ i1 ≤ i ^ i1, 0 < i1 ∧ i ^ i1 ≤ i0 ^ i1

Rule argument

Additional information

Used simplifier rules: notneg: ⊦ 0 ≤ i ∧ 0 ≤ j → ¬ i ^ j < 0;%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)