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Proof step 6 from proof gcd-mult (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ m ≤ n → gcd(m, n) = gcd(m, n - m)
Proof step 8:
n0 * m ≤ n0 * n, m < n, ∀ n1, m1, n2. n1 * m1 + n1 * n2 < n0 * m + n0 * n → true → gcd(n1 * m1, n1 * n2) = n1 * gcd(m1, n2) ⊦ n0 = 0, gcd(n0 * m, n0 * n - n0 * m) = n0 * gcd(m, n)

Rule argument

spec: inst: name: gcd-rec2 seq: ⊦ m ≤ n → gcd(m, n) = gcd(m, n - m) subst: [n, m] → [n0 * n, n0 * m] flags: rewrite, current

Additional information

Used simplifier rules: le: ⊦ m0 ≠ 0 → (m0 * m ≤ m0 * n ↔ m ≤ n); from specification nat-mult%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult Used terms: n0 * n, n0 * m