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Proof step 2 from proof gcd-le-both-h (previous step)

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) ∧ ¬ n < gcd(m, n)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ m ≤ n → gcd(m, n) = gcd(m, n - m)
Proof step 4:
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)

Rule argument

spec: nat-prime inst: name: gcd-rec2 seq: ⊦ m ≤ n → gcd(m, n) = gcd(m, n - m) subst: [n, m] → [n, m] flags: rewrite, current

Additional information

Used simplifier rules: lels-03: ⊦ m < n → m ≤ n; from specification nat Used terms: n, m