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Proof step 4 from proof l-02 (previous step)

m0 +1 + n < m0 +1, ∀ n. m0 + n < m0 → m0 = 0 ⊦ m0 +1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
m0 + n < m0 ⊦ m0 = 0, ∃ n. m0 + n < m0

Rule argument

Additional information

Used simplifier rules: disj: ⊦ 0 ≠ n +1;%rec: ⊦ m +1 + n = (m + n)+1;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%lrec: ⊦ m +1 < n +1 ↔ m < n;%f: ⊦ m < n → n ≠ 0;