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Proof step 9 from proof align-down-add-ls (previous step)

n0 = m0 * m + n2, (n0 + n1) / m * m < n0, n2 < m ⊦ (n0 + n1) % m = 0, m = 0, n2 + n1 < m ∧ n2 ≠ 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
n2 < m, ((n2 + n1) / m + m0) * m < m0 * m + n2, m ≤ n2 + n1 ⊦ (n2 + n1) % m = 0, n1 = 0, n2 = 0

Rule argument

Additional information

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