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Proof step 4 from proof gcd-trans (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n0 | n ⊦ gcd(m, n0) | gcd(m, n), gcd(m, n0) | n

Rule argument

Additional information

Used simplifier rules: gcd-divs: ⊦ gcd(m, n) | m;%gcd-a: ⊦ gcd(m, gcd(n, n0)) = gcd(gcd(m, n), n0);