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Proof step 28 from proof le-pos (previous step)

j0 < 0, ~2 < i0, 1 < i, j0 < i0, i ^ ~ j0 < i, j0 ≤ i0, 2 ≤ i, 0 ≤ i0, ~1 ≤ i0, i ^ ~ j0 ≤ 1, ∀ 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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 29:
⊦ 1 < i ∧ 0 < j → ¬ i ^ j ≤ 1

Rule argument

spec: <toplevel> name: notone seq: ⊦ 1 < i ∧ 0 < j → ¬ i ^ j ≤ 1 subst: [j, i] → [~ j0, i] flags: <all paths>

Additional information

Used simplifier rules: ⊦ 0 < ~ i ↔ i < 0; (s) Used terms: ~ j0, i