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Proof step 11 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) ⊦ m = n, n = 0, ∃ m0, n0. m * m0 + gcd(n, m % n) = n * n0 ∨ n * n0 + gcd(n, m % n) = m * m0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
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

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-02: ⊦ m ≠ n → (m ≤ n ↔ m < n); from specification nat