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Proof step 8 from proof ls-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
m1 * (n + n2) + n1 < m1 * n, n1 < n + n2, m < m1 * n, m1 * (n + n2) + n1 ≤ m ⊦ m1 = 0, n = 0

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → 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