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Proof step 6 from proof pot-square-est-01 (previous step)

4 ^ i ≤ i * i, 0 ≤ i, ∀ i0. i→n(i0) < i→n(i) → 4 ^ i0 ≤ i0 * i0 ∧ 0 ≤ i0 → false ⊦ 4 ^ (i - 1) ≤ (i - 1) * (i - 1) ∧ 0 < i

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
i = i1 + 1, 4 ^ i ≤ i * i, 0 ≤ i, ∀ i0. i→n(i0) < i→n(i) → 4 ^ i0 ≤ i0 * i0 ∧ 0 ≤ i0 → false ⊦ 4 ^ i1 ≤ i1 * i1 ∧ 0 < i

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [i, 1, i1] flags:

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i, 1, i1