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

m < gcd(m, n1), ∀ n0, m0. n0 < n1 + m → m0 < n0 → n0 = 0 ∨ m0 = 0 ∨ ¬ m0 < gcd(m0, n0) ∧ ¬ n0 < gcd(m0, n0) ⊦ m < n1, n1 + m = 0, m = 0, n1 = 0, gcd(m, n1) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
m < gcd(m, n1), n1 ≤ m, n1 < gcd(m, n1), ∀ n0, m0. n0 < n1 + m → m0 < n0 → n0 = 0 ∨ m0 = 0 ∨ ¬ m0 < gcd(m0, n0) ∧ ¬ n0 < gcd(m0, n0) ⊦ n1 + m = 0, m = 0, n1 = 0, gcd(m, n1) = 0

Rule argument

Additional information

Used simplifier rules: fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat