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Proof step 2 from proof gcd-max (previous step)

n0 | m, n0 | n ⊦ m = 0, n0 ≤ gcd(m, n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
gcd(m, n) < n0, n0 | m, n0 | n ⊦ m = 0

Rule argument

Additional information

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