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Proof step 3 from proof gcd-mult (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ n ≤ m → gcd(m, n) = gcd(m - n, n)
Proof step 5:
¬ n0 * n ≤ n0 * m, ∀ n1, m1, n2. n1 * m1 + n1 * n2 < n0 * m + n0 * n → true → gcd(n1 * m1, n1 * n2) = n1 * gcd(m1, n2) ⊦ gcd(n0 * m, n0 * n) = n0 * gcd(m, n)
Proof step 17:
n0 * n ≤ n0 * m, ∀ n1, m1, n2. n1 * m1 + n1 * n2 < n0 * m + n0 * n → true → gcd(n1 * m1, n1 * n2) = n1 * gcd(m1, n2) ⊦ gcd(n0 * m - n0 * n, n0 * n) = n0 * gcd(m, n)

Rule argument

spec: inst: name: gcd-rec1 seq: ⊦ n ≤ m → gcd(m, n) = gcd(m - n, n) subst: [n, m] → [n0 * n, n0 * m] flags: rewrite, current

Additional information

Used terms: n0 * n, n0 * m