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Proof step 4 from proof gcd-a (
previous step
)
⊦ gcd(m, gcd(n, n0)) | gcd(gcd(m, n), n0), gcd(m, gcd(n, n0)) = gcd(gcd(m, n), n0)
User applied rule weakening
And obtained the following new goals
Proof step 5
:
⊦ gcd(m, gcd(n, n0)) | gcd(gcd(m, n), n0)
Rule argument
poslist: right 2
Additional information