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Proof step 12 from proof gcdexh (previous step)

n < m, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ n = 0, m = n, ∃ m0, n0. m * m0 + gcd(n, m % n) = n * n0 ∨ n * n0 + gcd(n, m % n) = m * m0

User applied rule apply induction

And obtained the following new goals

Proof step 13:
∃ m1, n0. n * m1 + gcd(n, m % n) = m % n * n0 ∨ m % n * n0 + gcd(n, m % n) = n * m1, n < m, ∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ n = 0, m = n, ∃ m0, n0. m * m0 + gcd(n, m % n) = n * n0 ∨ n * n0 + gcd(n, m % n) = m * m0

Rule argument

subst: [m0, n] → [n, m % n]

Additional information

Used simplifier rules: mod-le-02: n ≠ 0 ⊦ m % n ≤ n; from specification nat-div Used terms: n, m % n