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Proof step 24 from proof le-pos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
j0 < 0, ~2 < i0, 1 < i, j0 < i0, j0 ≤ i0, 2 ≤ i, 0 ≤ i0, ~1 ≤ i0, ∀ j1, i, j0. i→n(j1) < i→n(i0 +1) → 1 < i ∧ 2 ≤ i ∧ 0 ≤ j1 → (i ^ j1 ≤ i ^ j0 ↔ j1 ≤ j0) ⊦ i ^ i0 * i ≤ 1 / i ^ ~ j0 ↔ i0 + 1 ≤ j0, j0 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ 0 - i = ~ i; (s)