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

m0 = m1 * n0 + n1, m < m1 * n, m0 < m1 * n, n1 < n0, m0 ≤ m, n ≤ n0 ⊦ m1 = 0, n = 0, n0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
m1 * n0 + n1 < m1 * n, m < m1 * n, n1 < n0, m1 * n0 + n1 ≤ m, n ≤ n0 ⊦ m1 = 0, n = 0

Rule argument

Additional information

Used simplifier rules: 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 ≠ 0; (s)