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Proof step 2 from proof ls-06 (previous step)

m < m0, n < n0 ⊦ m * n < m0 * n0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

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)