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Proof step 8 from proof gcd-le-both (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
n < m ⊦ m = n, n = 0, gcd(m, n) ≤ m ∧ gcd(m, n) ≤ n

Rule argument

Additional information

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