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Proof step 11 from proof gcd-div-both-h (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
n0 | gcd(m, n2), n2 ≤ m, ∀ n1, m0, n0. n1 < n2 + m → m0 < n1 ∧ n0 | gcd(m0, n1) → n0 | n1 ∧ n0 | m0 ⊦ n2 + m = 0, n2 = 0, n2 < m ∧ n0 | gcd(n2, m), n0 | n2 ∧ n0 | m

Proof step 11 from proof gcd-div-both-h (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information