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Proof step 10 from proof pot-square-est-02 (previous step)

0 < i1, 1 < i1, 0 ≤ i1, 4 ^ i1 * 4 ≤ (2 + i1) * (2 + 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
(i1 + 1) * (i1 + 1) < 4 ^ i1, 1 < i1, 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 ⊦

Rule argument

Additional information

Used simplifier rules: forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1% ⊦ i ≤ j ↔ ¬ j < i; (ss)