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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: succ: ⊦ m +1 = m + 1;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;