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Proof step 29 from proof le-pos (previous step)

1 < i ^ ~ j0, 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 ≤ 0 ↔ i0 + 1 ≤ j0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 30:
2 ≤ i ^ ~ j0, j0 < i0, j0 < 0, ~1 ≤ i0, 0 ≤ i0, ~2 < i0, 1 < i, 2 ≤ i, j0 ≤ i0, 1 < i ^ ~ j0, ∀ j1, i, j0. i→n(j1) < i→n(i0 +1) → 2 ≤ i ∧ 0 ≤ j1 ∧ 1 < i → (i ^ j1 ≤ i ^ j0 ↔ j1 ≤ j0) ⊦ i ^ i0 * i ≤ 0 ↔ i0 + 1 ≤ j0, j0 = 0

Rule argument

Additional information

Used simplifier rules: numlef-01: ⊦ (* i) ≤ j → i - 1 < j; (forward) from specification int-abs%numlsf-02: ⊦ (* i) < j → i + 1 ≤ j; (forward) from specification int-abs