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

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

User applied rule cut formula

And obtained the following new goals

Proof step 9:
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

Rule argument

fma: 1 < i1

Additional information

Used simplifier rules: lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%bet: ⊦ 0 < i → (i ≤ 1 ↔ i = 1); from specification int-abs%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used formulas: 1 < i1