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Proof step 15 from proof gcd-a (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 16:
⊦ gcd(gcd(n, m), n0) | m ∧ gcd(gcd(n, m), n0) | gcd(n, n0), gcd(gcd(n, m), n0) | gcd(m, gcd(n, n0))

Rule argument

pos: left 1

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