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Proof step 22 from proof gcd-a (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
⊦ gcd(n0, gcd(m, n)) | gcd(m, gcd(n, n0)), gcd(n0, gcd(m, n)) | gcd(n, n0)

Rule argument

Additional information

Used simplifier rules: gcd-div-03: ⊦ gcd(m, gcd(n, n0)) | n;