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Proof step 4 from proof gcdex (previous step)

m < n ⊦ ∃ m0, n0. m * m0 + gcd(m, n) = n * n0 ∨ n * n0 + gcd(m, n) = m * m0

User applied rule insert lemma

And obtained the following new goals

Proof step 5:
n ≤ m ⊦ ∃ m0, n0. m * m0 + gcd(m, n) = n * n0 ∨ n * n0 + gcd(m, n) = m * m0
Proof step 6:
m < n, ∃ m0, n0. n * m0 + gcd(n, m) = m * n0 ∨ m * n0 + gcd(n, m) = n * m0 ⊦ ∃ m0, n0. m * m0 + gcd(m, n) = n * n0 ∨ n * n0 + gcd(m, n) = m * m0

Rule argument

spec: <toplevel> name: gcdexh seq: n ≤ m ⊦ ∃ m0, n0. m * m0 + gcd(m, n) = n * n0 ∨ n * n0 + gcd(m, n) = m * m0 subst: [n, m] → [m, n]

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < n → ¬ n < m; (s) Used terms: m, n