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Proof step 10 from proof gcd-mod (previous step)

n0 < n, ∀ n0, n. n0 < n → gcd(m * n + n0, n) = gcd(n, n0) ⊦ gcd((n * m + n) + n0, n) = gcd(n, n0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ n ≤ m → gcd(m, n) = gcd(m - n, n)
Proof step 12:
n0 < n, n ≤ (n * m + n) + n0, ∀ n0, n. n0 < n → gcd(m * n + n0, n) = gcd(n, n0) ⊦ gcd((n * m + n) + n0 - n, n) = gcd(n, n0)

Rule argument

spec: <toplevel> name: gcd-rec1 seq: ⊦ n ≤ m → gcd(m, n) = gcd(m - n, n) subst: [n, m] → [n, (n * m + n) + n0] flags: <all paths>

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: n, (n * m + n) + n0