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Proof step 9 from proof pot-square-est-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
4 ^ i1 * 4 ≤ (2 + i1) * (2 + i1), 0 ≤ i1, (i1 + 1) * (i1 + 1) < 4 ^ i1, 1 < i1, 0 < i1, ∀ i0. i→n(i0) < i→n(i1 + 1) → 4 ^ i0 ≤ (i0 + 1) * (i0 + 1) ∧ 0 < i0 ∧ 1 < i0 → false ⊦

Rule argument

Additional information

Used simplifier rules: forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs