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Proof step 5 from proof gcd-le-both (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat