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Proof step 3 from proof l-05 (previous step)

m + m0 < n, n + n0 < m ⊦ m = 0, n = 0

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 4:
n + n0 < m, m + m1 +1 < n, ∀ m, n, n0. m + m1 < n ∧ n + n0 < m → m = 0 ∨ n = 0 ⊦ m = 0, n = 0

Rule argument

term: m0

Additional information

Used simplifier rules: gen: ⊦ ;%z-02: ⊦ 0 + m = m;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%trans-01: ⊦ n < n0 ∧ n0 < n1 → n +1 < n1;%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1;%succ: ⊦ m +1 = m + 1;%l-02: ⊦ ¬ m + n < m;