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Proof step 4 from proof even-06 (previous step)

m + m = 3 + n0 + n0, ∀ m. m + m ≠ n0 + n0 + 1 ⊦

User applied rule structural induction

And obtained the following new goals

Proof step 5:
∀ 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 ⊦

Rule argument

indvar: m

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0;%nat freely generated by 0, +1;;