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Proof step 1 from proof one-01 (conclusion)

⊦ m ≠ 0 → (m + n ≤ 1 ↔ m = 1 ∧ n = 0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
m + n ≤ 1 ⊦ m = 0, m = 1 ∧ n = 0

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2