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Proof step 2 from proof gcd-div-both-h (previous step)

m < n ∧ n0 | gcd(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 3:
m < n, n0 | gcd(m, n), ∀ n1, m, n0. n1 < n → m < n1 ∧ n0 | gcd(m, n1) → n0 | n1 ∧ n0 | m ⊦ n = 0, n0 | n ∧ n0 | m

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1