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Proof step 3 from proof gcdexh (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ m0, n. m0 < m → n ≤ m0 → (∃ m1, n0. m0 * m1 + gcd(m0, n) = n * n0 ∨ n * n0 + gcd(m0, n) = m0 * m1) ⊦ ∃ m0. m * m0 + m = 0 ∨ m = 0 ∨ m0 = 1

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);%eq-same: ⊦ m * n = m ↔ ¬ ¬(m = 0 ∨ n = 1); from specification nat-mult%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%z: ⊦ 0 + m = m; 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%mult-base: ⊦ m * 0 = 0; from specification nat-mult%lez: ⊦ 0 ≤ m; from specification nat