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Proof step 3 from proof l-03 (previous step)

n < n0, m < m0 ⊦ m0 = 0, n0 = 0, m + n < m0 + n0

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 4:
m < m0, n < n1 +1, ∀ n, m, m0. n < n1 ∧ m < m0 → m0 = 0 ∨ n1 = 0 ∨ m + n < m0 + n1 ⊦ n1 +1 = 0, m0 = 0, m + n < m0 + n1 +1

Rule argument

term: n0

Additional information

Used simplifier rules: gen: ⊦ ;