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Proof step 4 from proof l (previous step)

m + n < n1 + m, ∀ m, n. m + n < m + n1 ↔ n < n1 ⊦ m + n < m + n1 + 1 ↔ n < n1 + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
m + n < n1 + m, n1 + 1 < n, ∀ m, n. m + n < m + n1 ↔ n < n1 ⊦ n1 + m = 0, n = n1 + 1, n = 0

Rule argument

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%l-08: ⊦ m + n < m0 → m + n < m0 + n0;%ls-04: ⊦ m < n ↔ ¬(m = n ∨ n < m);%f: ⊦ m < n → n ≠ 0;%l-02: ⊦ ¬ m + n < m;