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

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

User applied rule cut formula

And obtained the following new goals

Proof step 10:
n0 + n < (n0 / n + 1) * n + n, n0 < n * n0 / n + n, n0 / n * n ≤ n0, n0 / n + 1 < (n0 + n) / n, (n0 + n) / n * n ≤ n0 + n, n0 < n * (n0 + n) / n, n0 / n < (n0 + n) / n ⊦ (n0 + n) / n * n < (n0 / n + 1) * n + n, n * n0 / n + n = 0, n = 0, (n0 + n) / n = 0

Rule argument

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

Additional information

Used simplifier rules: rsucc-02: ⊦ m * (n + 1) = m * n + m; from specification nat-mult%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%two-02: ⊦ m + m = 2 * m; from specification nat-mult%numeral-24: m1 ≠ 0 ⊦ m + n < (* m1) * m + n0 ↔ n < (m1 - 1) * m + n0; from specification nat-mult%one: ⊦ m0 * 1 = m0; from specification nat-mult%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat Used formulas: n0 + n < (n0 / n + 1) * n + n