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

gcd(m, n) = m ⊦ n = 0, m ≤ n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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