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Proof step 3 from proof gcd-le-both (previous step)

⊦ m = 0, n = 0, gcd(m, n) ≤ m ∧ gcd(m, n) ≤ n

User applied rule cut formula

And obtained the following new goals

Proof step 4:
m < n ⊦ m = 0, n = 0, gcd(m, n) ≤ m ∧ gcd(m, n) ≤ n
Proof step 6:
⊦ m = 0, n = 0, m < n, gcd(m, n) ≤ m ∧ gcd(m, n) ≤ n

Rule argument

fma: m < n

Additional information

Used formulas: m < n