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Proof step 3 from proof gcd-div-both-h (previous step)

m < n, n0 | gcd(m, n), ∀ n1, m, n0. n1 < n → m < n1 ∧ n0 | gcd(m, n1) → n0 | n1 ∧ n0 | m ⊦ n0 | n ∧ n0 | m

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ m ≤ n → gcd(m, n) = gcd(m, n - m)
Proof step 5:
m < n, n0 | gcd(m, n - m), m ≤ n, ∀ n1, m, n0. n1 < n → m < n1 ∧ n0 | gcd(m, n1) → n0 | n1 ∧ n0 | m ⊦ n0 | n ∧ n0 | m

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