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Proof step 6 from proof gcd-le-both (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n ≤ m ⊦ m = 0, n = 0, gcd(m, n) ≤ m ∧ gcd(m, n) ≤ n

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)