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Proof step 10 from proof square-ls-04 (previous step)

m * m + m0 < (m + 1) * (m + 1), m < n, m0 ≤ 2 * m, n * n ≤ m * m + m0 ⊦ n = 0

User applied rule cut formula

And obtained the following new goals

Proof step 11:
n * n < (m + 1) * (m + 1), m * m + m0 < (m + 1) * (m + 1), m < n, m0 ≤ 2 * m, n * n ≤ m * m + m0 ⊦ n = 0

Rule argument

fma: n * n < (m + 1) * (m + 1)

Additional information

Used simplifier rules: rsucc-01: ⊦ (m + 1) * n = m * n + n;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%two-02: ⊦ m + m = 2 * m;%zero-06: ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0);%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat Used formulas: n * n < (m + 1) * (m + 1)