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Proof step 5 from proof even-02 (previous step)

2 + m0 + m0 < n + n, n < m0, m0 + m0 < n -1 + n -1 ∧ n -1 < m0 → n -1 + n -1 = 0 ∨ m0 = n -1 ⊦ m0 + 1 = n, n + n = 0, m0 = n, m0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n < m0, 2 + m0 + m0 < n + n, m0 + m0 < n -1 + n -1 → n -1 + n -1 = 0 ∨ m0 = n -1 ⊦ n + n = 0, m0 + 1 = n, m0 = 0, m0 = n

Rule argument

Additional information

Used simplifier rules: oz: ⊦ m + 1 ≠ 0;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%psh-09: m ≠ 0 ⊦ m -1 < n ↔ m < n +1;%succ: ⊦ m +1 = m + 1;%neq-04: ⊦ m ≠ n → (m < n + 1 ↔ m < n);