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Proof step 1 from proof l-03 (conclusion)

⊦ m < m0 ∧ n < n0 → (¬ m + n < m0 + n0 ↔ false)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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