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Proof step 11 from proof pot-square-est-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
i1 * i1 < 4 ^ i1, 4 ^ i1 * 4 ≤ (i1 + 1) * (i1 + 1), 0 ≤ i1, ∀ i0. i→n(i0) < i→n(i1 + 1) → 4 ^ i0 ≤ i0 * i0 ∧ 0 ≤ i0 → false ⊦

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j; (s)