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Proof step 4 from proof two-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
1 + n = 2 ⊦ n = 1, n = 0

Rule argument

Additional information

Used simplifier rules: succ: ⊦ m +1 = m + 1;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%oz: ⊦ m + 1 ≠ 0;%c: ⊦ m + n = n + m;%one-03: ⊦ 1 = m + 1 ↔ m = 0;%one-01: ⊦ n ≠ 0 → (m + n = 1 ↔ n = 1 ∧ m = 0);