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Proof step 3 from proof even-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n < m0, 2 + m0 + m0 < n + n, ∀ n0. m0 + m0 < n0 + n0 ∧ n0 < m0 → n0 + n0 = 0 ∨ m0 = n0 ⊦ n + n = 0, m0 + 1 = n, m0 = 0, m0 = n

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-04: ⊦ m < n + 1 ↔ ¬ n < m;%ls-04: ⊦ m < n ↔ ¬(m = n ∨ n < m);%f: ⊦ m < n → n ≠ 0;%eq-02: ⊦ m = m + n ↔ n = 0;%l: ⊦ m + n < m + n0 ↔ n < n0;%l-02: ⊦ ¬ m + n < m;