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Proof step 8 from proof gcd-zero-01-h (previous step)

n = n1 + m ∧ m ≤ n, gcd(m, n1) = 0, m < n, ∀ n0, m. n0 < n → gcd(m, n0) = 0 ∧ m < n0 → n0 = 0 ⊦ n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
gcd(m, n1) = 0, ∀ n0, m0. n0 < n1 + m → gcd(m0, n0) = 0 ∧ m0 < n0 → n0 = 0 ⊦ n1 + m = 0, n1 = 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%l-01: ⊦ m < m + n ↔ n ≠ 0; from specification nat-basic2%lels-03: ⊦ m < n → m ≤ n; from specification nat