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

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 ^ ~ j0 ↔ i0 + 1 ≤ j0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 25:
⊦ 1 < j → 1 / j = 0
Proof step 26:
¬ 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 ≤ 1 / i ^ ~ j0 ↔ i0 + 1 ≤ j0
Proof step 29:
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

Rule argument

spec: int-div inst: name: one-04 seq: ⊦ 1 < j → 1 / j = 0 subst: [j] → [i ^ ~ j0] flags: rewrite

Additional information

Used terms: i ^ ~ j0