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Proof step 5 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 - 1) + 1) * ((i - 1) + 1) ∧ 0 < i - 1 ∧ 1 < i - 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
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

Rule argument

Additional information

Used simplifier rules: numlelsdel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) < j → (* i0) < j; from specification int-abs%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j; 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%Sub-def: ⊦ i + (j - i) = j; from specification int-abs