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Proof step 10 from proof square-ls-05 (previous step)

m * m + m * m0 < n * n + m * m0, m < n, n * n + n * m0 ≤ m * m + m * m0 ⊦ n = 0

User applied rule cut formula

And obtained the following new goals

Proof step 11:
n * n + m * m0 < n * n + n * m0, m * m + m * m0 < n * n + m * m0, m < n, n * n + n * m0 ≤ m * m + m * m0 ⊦ n = 0

Rule argument

fma: n * n + m * m0 < n * n + n * m0

Additional information

Used simplifier rules: ls-06: ⊦ m < m0 ∧ n < n0 → m * n < m0 * n0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%ls: ⊦ m0 * m < m0 * n ↔ m0 ≠ 0 ∧ m < n;%zero: ⊦ 0 * n = 0;%square-le-02: ⊦ m * m ≤ n * n ↔ m ≤ n;%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%add-base: ⊦ n + 0 = n; from specification nat-basic2%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat Used formulas: n * n + m * m0 < n * n + n * m0