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Proof step 3 from proof gcd-a (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 4:
⊦ gcd(m, gcd(n, n0)) | gcd(gcd(m, n), n0), gcd(m, gcd(n, n0)) = gcd(gcd(m, n), n0)
Proof step 10:
⊦ gcd(gcd(m, n), n0) | gcd(m, gcd(n, n0)), gcd(m, gcd(n, n0)) = gcd(gcd(m, n), n0)

Rule argument

pos: right 1

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