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Proof step 9 from proof gcd-alldiv-h (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
n0 | m, n0 | n2, n2 ≤ m, ∀ n1, m0, n0. n1 < n2 + m → m0 < n1 ∧ n0 | n1 ∧ n0 | m0 → n0 | gcd(m0, n1) ⊦ n2 + m = 0, n2 = 0, n0 | gcd(m, n2)

Rule argument

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat