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Proof step 7 from proof gcd-le-both-h (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
∀ n0, m0. n0 < n1 + m → m0 < n0 → m0 = 0 ∨ n0 = 0 ∨ ¬ m0 < gcd(m0, n0) ∧ ¬ n0 < gcd(m0, n0) ⊦ m = 0, n1 = 0, ¬ m < gcd(m, n1) ∧ ¬ n1 + m < gcd(m, n1)

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ m < n → m ≤ n; (s)