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Proof step 9 from proof gcdexh (previous step)

m = n, 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, ∃ 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 10:
∀ m0, n0. m0 < n → n0 ≤ m0 → (∃ m1, n1. m0 * m1 + gcd(m0, n0) = n0 * n1 ∨ n0 * n1 + gcd(m0, n0) = m0 * m1) ⊦ n = 0, ∃ m, n0. m + 1 = n0 ∨ n0 + 1 = m

Rule argument

Additional information

Used simplifier rules: gcd-zero-02: ⊦ gcd(n, 0) = n;%gcd-a: ⊦ gcd(m, gcd(n, n0)) = gcd(gcd(m, n), n0);%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%coef-17: ⊦ m * n + n = m1 * n ↔ ¬ ¬(m + 1 = m1 ∨ n = 0); from specification nat-mult%mod-zero: n ≠ 0 ⊦ n % n = 0; from specification nat-div%ler: ⊦ m ≤ m; from specification nat