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Proof step 3 from proof gcd-div-03 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ gcd(m, gcd(n, n0)) | n, gcd(m, gcd(n, n0)) | gcd(n, n0)

Rule argument

Additional information

Used simplifier rules: gcd-div: ⊦ n0 | gcd(m, n) → n0 | m;%gcd-div-01: ⊦ n0 | gcd(m, n) → n0 | n;