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Proof step 1 from proof gcd-max (conclusion)

⊦ m ≠ 0 ∧ n0 | m ∧ n0 | n → n0 ≤ gcd(m, n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n0 | m, n0 | n, gcd(m, n) < n0 ⊦ m = 0, n0 = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat