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Proof step 3 from proof twols (previous step)

∀ m. m < n0 ∧ m + n0 < 2 → n0 = 0 ∨ m = 0 ∧ n0 = 1, m < n0 +1, m + n0 +1 < 2 ⊦ n0 +1 = 0, m = 0 ∧ n0 +1 = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
m + n0 + 1 < 2, ∀ m. m < n0 ∧ m + n0 < 2 → n0 = 0 ∨ m = 0 ∧ n0 = 1 ⊦ n0 < m, m = 0 ∧ n0 = 0

Rule argument

Additional information

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