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Proof step 22 from proof le-pos (previous step)

~2 < i0, 1 < i, j0 < 0, 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 ^ abs(j0) ↔ i0 + 1 ≤ j0, j0 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 23:
⊦ j < 0 → abs(j) = 0 - j
Proof step 24:
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

Rule argument

spec: int-abs name: Abs-neg seq: ⊦ j < 0 → abs(j) = 0 - j subst: [j] → [j0] flags: <all paths>

Additional information

Used terms: j0