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Proof step 1 from proof le-add-02 (conclusion)

⊦ n < m ∧ n0 < m0 → (¬ ¬ m + m0 ≤ n + n0 + 1 ↔ false)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n < m, n0 < m0, m + m0 ≤ n + n0 + 1 ⊦ m = 0, m0 = 0

Rule argument

Additional information

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