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Proof step 5 from proof le-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n0 < n * (n0 + n) / n, (n0 + n) / n * n ≤ n0 + n, (n0 + n) / n ≤ n0 / n ⊦ (n0 + n) / n = 0, n = 0

Rule argument

Additional information

Used simplifier rules: ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0); (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m * (n + 1) = m * n + m; (s)% ⊦ m +1 = m + 1; (s)