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Proof step 9 from proof gcd-le-both-h (previous step)

m < gcd(m, n1), n1 ≤ m, n1 < gcd(m, n1), ∀ n0, m0. n0 < n1 + m → m0 < n0 → n0 = 0 ∨ m0 = 0 ∨ ¬ m0 < gcd(m0, n0) ∧ ¬ n0 < gcd(m0, n0) ⊦ n1 + m = 0, m = 0, n1 = 0, gcd(m, n1) = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ gcd(m, n) = gcd(n, m)
Proof step 11:
m < gcd(n1, m), n1 ≤ m, n1 < gcd(n1, m), ∀ n0, m0. n0 < n1 + m → m0 < n0 → n0 = 0 ∨ m0 = 0 ∨ ¬ m0 < gcd(m0, n0) ∧ ¬ n0 < gcd(m0, n0) ⊦ n1 + m = 0, m = 0, n1 = 0, gcd(n1, m) = 0

Rule argument

spec: nat-prime inst: name: gcd-c seq: ⊦ gcd(m, n) = gcd(n, m) subst: [n, m] → [n1, m] flags: rewrite, current

Additional information

Used terms: n1, m