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Proof step 8 from proof pot-square-est-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
0 < i1, 4 ^ i1 * 4 ≤ (2 + i1) * (2 + i1), 0 ≤ i1, ∀ i0. i→n(i0) < i→n(i1 + 1) → 0 < i0 ∧ 1 < i0 ∧ 4 ^ i0 ≤ (i0 + 1) * (i0 + 1) → false ⊦ 1 < i1 ∧ 4 ^ i1 ≤ (i1 + 1) * (i1 + 1)

Rule argument

Additional information

Used simplifier rules: succ: ⊦ 0 ≤ i → j ^ (i + 1) = j ^ i * j;%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs% ⊦ ~1 < i ↔ 0 ≤ i; (s)% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2