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Proof step 1 from proof gcd-eq (conclusion)

⊦ n ≠ 0 ∧ gcd(m, n) = m → m ≤ n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat