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Proof step 7 from proof pot-square-est-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
4 ^ (i1 + 1) ≤ (i1 + 1) * (i1 + 1), ~1 ≤ i1, ∀ i0. i→n(i0) < i→n(i1 + 1) → 4 ^ i0 ≤ i0 * i0 ∧ 0 ≤ i0 → false ⊦ 4 ^ i1 ≤ i1 * i1 ∧ 0 ≤ i1

Rule argument

Additional information

Used simplifier rules: mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs