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Proof step 7 from proof gcd-div-both-h (previous step)

n = n2 + m, m < n, n0 | gcd(m, n2), m ≤ n, ∀ n1, m, n0. n1 < n → m < n1 ∧ n0 | gcd(m, n1) → n0 | n1 ∧ n0 | m ⊦ n0 | n ∧ n0 | m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
n0 | gcd(m, n2), ∀ n1, m0, n0. n1 < n2 + m → m0 < n1 ∧ n0 | gcd(m0, n1) → n0 | n1 ∧ n0 | m0 ⊦ n2 = 0, n0 | n2 ∧ n0 | m

Rule argument

Additional information

Used simplifier rules: divi-add: ⊦ m | n → (m | n + n0 ↔ m | n0);%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ m < n → m ≤ n; (s)