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Proof step 6 from proof pot-square-est-02 (previous step)

4 ^ i ≤ (i + 1) * (i + 1), 0 < i, 1 < i, ∀ i0. i→n(i0) < i→n(i) → 4 ^ i0 ≤ (i0 + 1) * (i0 + 1) ∧ 0 < i0 ∧ 1 < i0 → false ⊦ 4 ^ (i - 1) ≤ i * i ∧ 2 < i

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
i = i1 + 1, 4 ^ i ≤ (i + 1) * (i + 1), 0 < i, 1 < i, ∀ i0. i→n(i0) < i→n(i) → 4 ^ i0 ≤ (i0 + 1) * (i0 + 1) ∧ 0 < i0 ∧ 1 < i0 → false ⊦ 4 ^ i1 ≤ i * i ∧ 2 < 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