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Proof step 1 from proof ls-add-not (conclusion)

⊦ m ≤ m0 ∧ n ≤ n0 → ¬ m0 + n0 + n1 < m + n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
m0 + n0 + n1 < m + n, m ≤ m0, n ≤ n0 ⊦ m + n = 0

Rule argument

Additional information

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