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Proof step 4 from proof pot-square-est-02 (previous step)

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

User applied rule apply induction

And obtained the following new goals

Rule argument

subst: [i0] → [i - 1]

Additional information

Used simplifier rules: rax-01: ⊦ 1 < i → i→n(i - 1) = i→n(i) -1; from specification intnat%pred-03: ⊦ n -1 = n - 1; from specification nat%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%num-i2n-le-02: 0 ≤ i ⊦ (* m) ≤ i→n(i) ↔ n→i(m) ≤ i; from specification intnat%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs%ls-01: n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; from specification nat%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%labs: ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); from specification intnat%abs: ⊦ 1 < i → abs(i) = i; from specification int-abs%irref: ⊦ ¬ i < i; from specification int-basic1%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: i - 1