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

n0 < m / n ⊦ n = 0, n0 * n < m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n ↔ ¬ n ≤ m; (ss)