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Proof step 9 from proof ls-03 (previous step)

n0 < n * n0 / n + n, n0 / n + 1 < (n0 + n) / n, n0 < n * (n0 + n) / n, n0 / n < (n0 + n) / n, n0 / n * n ≤ n0, (n0 + n) / n * n ≤ n0 + n ⊦ n = 0

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: (n0 + n) / n * n < (n0 / n + 1) * n + n

Additional information

Used simplifier rules: ls-02: n ≠ 0 ⊦ m / n < m0 / n → m < m0;% ⊦ m * (n + 1) = m * n + m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ m + m = 2 * m; (s)% ⊦ m * n < m1 * n + m2 * n ↔ m < m1 + m2 ∧ n ≠ 0; (s)% ⊦ m < n → (n < m + 2 ↔ m + 1 = n); (s)%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic% ⊦ m < n → (n ≤ m ↔ false); (s) Used formulas: (n0 + n) / n * n < (n0 / n + 1) * n + n