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Proof step 3 from proof divi-primmult (previous step)

prime(m), m | n * n0 ⊦ m | n, m | n0

User applied rule cut formula

And obtained the following new goals

Proof step 4:
gcd(m, n) = 1, prime(m), m | n * n0 ⊦ m | n0, m | n
Proof step 6:
prime(m), m | n * n0 ⊦ gcd(m, n) = 1, m | n0, m | n

Rule argument

fma: gcd(m, n) = 1

Additional information

Used formulas: gcd(m, n) = 1