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Proof step 3 from proof gcd-le-both-h (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ m ≤ n → gcd(m, n) = gcd(m, n - m)
Proof step 5:
m < n, m ≤ n, ∀ n0, m. n0 < n → m < n0 → m = 0 ∨ n0 = 0 ∨ ¬ m < gcd(m, n0) ∧ ¬ n0 < gcd(m, n0) ⊦ m = 0, n = 0, ¬ m < gcd(m, n - m) ∧ ¬ n < gcd(m, n - m)

Rule argument

spec: <toplevel> name: gcd-rec2 seq: ⊦ m ≤ n → gcd(m, n) = gcd(m, n - m) subst: [n, m] → [n, m] flags: <all paths>

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < n → ¬ n < m; (s) Used terms: n, m