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Proof step 5 from proof even-06 (previous step)

∀ n0. ¬(m0 + m0 = 3 + n0 + n0 ∧ (∀ m. m + m ≠ n0 + n0 + 1)), m0 +1 + m0 +1 = 3 + n0 + n0, ∀ m. m + m ≠ n0 + n0 + 1 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
2 + m0 + m0 = 3 + n0 + n0, ∀ m. m + m ≠ n0 + n0 + 1, ∀ n. m0 + m0 ≠ 3 + n + n ∨ ¬(∀ m. m + m ≠ n + n + 1) ⊦

Rule argument

Additional information

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