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Proof step 2 from proof ls-01 (previous step)

n ≤ n0, m0 ≤ m ⊦ n = 0, m0 / n0 ≤ m / n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
m0 < m0 / n0 * n, m < m0 / n0 * n, n ≤ n0, m0 ≤ m ⊦ m0 / n0 = 0, n = 0

Rule argument

Additional information

Used simplifier rules: le-div-to-mult: m0 ≠ 0 ⊦ n ≤ m / m0 ↔ n * m0 ≤ m;%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0); (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)