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Proof step 1 from proof gcd-div-both-h (conclusion)

m < n, n0 | gcd(m, n) ⊦ n0 | n ∧ n0 | m

User applied rule induction

And obtained the following new goals

Proof step 2:
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

Rule argument

indvar: n wfpred: <

Additional information

Used simplifier rules: <; (wellfounded) Used terms: n