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Proof step 8 from proof gcd-alldiv (previous step)

n0 | n, n0 | m ⊦ m < n, n0 | gcd(n, m)

User applied rule insert lemma

And obtained the following new goals

Proof step 9:
m < n, n0 | n, n0 | m ⊦ n0 | gcd(m, n)

Rule argument

spec: <toplevel> name: gcd-alldiv-h seq: m < n, n0 | n, n0 | m ⊦ n0 | gcd(m, n) subst: [m, n, n0] → [n, m, n0]

Additional information

Used simplifier rules: gcd-id: ⊦ gcd(m, m) = m;% ⊦ ¬ m < n → (n < m ↔ n ≠ m); (ss)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m ≤ m; (s) Used terms: n, m, n0