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Proof step 9 from proof gcd-div-both-h (previous step)

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, m < n2 ∧ n0 | gcd(m, n2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
n0 | gcd(m, n2), n2 ≤ m, ∀ 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: ⊦ m < n ↔ ¬ n ≤ m; (ss)