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Proof step 2 from proof gcd-div-both (previous step)

n0 | gcd(m, n) ⊦ n0 | n ∧ n0 | m

User applied rule cut formula

And obtained the following new goals

Proof step 3:
m < n, n0 | gcd(m, n) ⊦ n0 | n ∧ n0 | m
Proof step 5:
n0 | gcd(m, n) ⊦ m < n, n0 | n ∧ n0 | m

Rule argument

fma: m < n

Additional information

Used formulas: m < n