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Proof step 4 from proof gcd-le-both-h (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
m < n, ∀ n0, m. n0 < n → m < n0 → n0 = 0 ∨ m = 0 ∨ ¬ m < gcd(m, n0) ∧ ¬ n0 < gcd(m, n0) ⊦ n = 0, m = 0, ¬ m < gcd(m, n - m) ∧ ¬ n < gcd(m, n - m)

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-03: ⊦ m < n → m ≤ n; from specification nat