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Proof step 10 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 = 0, n0 | n2 ∧ n0 | m

User applied rule apply induction

And obtained the following new goals

Proof step 11:
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, n2 < m ∧ n0 | gcd(n2, m)

Rule argument

subst: [n1, m0, n0] → [m, n2, n0]

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m + m0 ≤ m ↔ m0 = 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: m, n2, n0