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Proof step 1 from proof ls-06 (conclusion)

⊦ m < m0 ∧ n < n0 → m * n < m0 * n0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
m < m0, n < n0, m0 * n0 ≤ m * n ⊦ m0 = 0, n0 = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat