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Proof step 6 from proof l (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
m + n1 + 1 < m + n, n < n1 + 1, n1 + m < m + n, ∀ m, n. m + n < m + n1 ↔ n < n1 ⊦ m + n = 0, n = n1 + 1, n = n1

Rule argument

Additional information

Used simplifier rules: ls-04: ⊦ m < n ↔ ¬(m = n ∨ n < m);%eq-01: ⊦ m + n = m + n0 ↔ n = n0;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%l-addone: ⊦ n < n + 1;%f: ⊦ m < n → n ≠ 0;%trans-01: ⊦ n < n0 ∧ n0 < n1 → n +1 < n1;%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1;%oz: ⊦ m + 1 ≠ 0;%succ: ⊦ m +1 = m + 1;%ref: ⊦ ¬ n < n;%eq-02: ⊦ m = m + n ↔ n = 0;