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Proof step 10 from proof gcd-le-both-h (previous step)

n1 ≤ m, ∀ 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)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ gcd(m, n) = gcd(n, m)
Proof step 12:
n1 ≤ m, ∀ 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(n1, m) ∧ ¬ n1 + m < gcd(n1, m)

Rule argument

spec: <toplevel> name: gcd-c seq: ⊦ gcd(m, n) = gcd(n, m) subst: [n, m] → [n1, m] flags: <all paths>

Additional information

Used terms: n1, m