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Proof step 11 from proof gcd-mult (previous step)

n0 * n = n3 + n0 * m, m < n, ∀ n1, m1, n2. n1 * m1 + n1 * n2 < n0 * m + n0 * n → true → gcd(n1 * m1, n1 * n2) = n1 * gcd(m1, n2) ⊦ gcd(n0 * m, n3) = n0 * gcd(m, n), n0 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ m ≤ n → gcd(m, n) = gcd(m, n - m)
Proof step 13:
n0 * n = n3 + n0 * m, m < n, m ≤ n, ∀ n1, m1, n2. n1 * m1 + n1 * n2 < n0 * m + n0 * n → true → gcd(n1 * m1, n1 * n2) = n1 * gcd(m1, n2) ⊦ gcd(n0 * m, n3) = n0 * gcd(m, n - m), n0 = 0

Rule argument

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

Additional information

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