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Proof step 3 from proof addex (previous step)

∀ m. m < n0 → n0 = 0 ∨ (∃ m0. m + m0 = n0 ∧ m0 ≠ 0), m < n0 +1 ⊦ n0 +1 = 0, ∃ m0. m + m0 = n0 +1 ∧ m0 ≠ 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ m. m < n0 → n0 = 0 ∨ (∃ m0. m + m0 = n0 ∧ m0 ≠ 0) ⊦ n0 < m, ∃ m0. m + m0 = n0 + 1 ∧ m0 ≠ 0

Rule argument

Additional information

Used simplifier rules: succ: ⊦ m +1 = m + 1;%oz: ⊦ m + 1 ≠ 0;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%s: ⊦ m < n + 1 ↔ ¬ n < m;