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Proof step 9 from proof l (previous step)

∀ m, n1, n2. ¬ m0 < m → (m + n1 < m + n2 ↔ n1 < n2) ⊦ n < n0, m0 + n < m0 + n0 ↔ n < n0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
m0 + n < m0 + n0, ∀ m, n1, n2. ¬ m0 < m → (m + n1 < m + n2 ↔ n1 < n2) ⊦ m0 + n0 = 0, n < n0

Rule argument

Additional information

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