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Proof step 5 from proof ls-02 (previous step)

m / n < m0 / n, m0 ≤ m, n ≠ 0 → m0 / n * n ≤ m0 ∧ m0 < (m0 / n)+1 * n ⊦ n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
m0 < n * m0 / n + n, m / n < m0 / n, m0 / n * n ≤ m, m0 / n * n ≤ m0, m0 ≤ m ⊦ n = 0

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < n; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m * n < m1 * n + n ↔ m < m1 + 1 ∧ n ≠ 0; (s)% ⊦ m < n → n ≠ 0; (s)%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) 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%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)