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Proof step 5 from proof gcd-zero-01-h (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ m ≤ n → gcd(m, n) = gcd(m, n - m)
Proof step 7:
gcd(m, n - m) = 0, m < n, m ≤ n, ∀ n0, m. n0 < n → gcd(m, n0) = 0 ∧ m < n0 → false ⊦

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 < m; (ss)% ⊦ m < n → ¬ n < m; (s) Used terms: n, m