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Proof step 10 from proof gcd-mult (previous step)

n0 * n = n3 + n0 * m ∧ n0 * m ≤ n0 * n, m < n, ∀ n1, m1, n2. n1 * m1 + n1 * n2 < n0 * m + n0 * n → true → gcd(n1 * m1, n1 * n2) = n1 * gcd(m1, n2) ⊦ n0 = 0, gcd(n0 * m, n3) = n0 * gcd(m, n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
n0 * n = n3 + n0 * m, m < n, ∀ n1, m1, n2. n1 * m1 + n1 * n2 < n0 * m + n0 * n → true → gcd(n1 * m1, n1 * n2) = n1 * gcd(m1, n2) ⊦ n0 = 0, gcd(n0 * m, n3) = n0 * gcd(m, n)

Rule argument

Additional information

Used simplifier rules: mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%lels-03: ⊦ m < n → m ≤ n; from specification nat%le: ⊦ m0 ≠ 0 → (m0 * m ≤ m0 * n ↔ m ≤ n); from specification nat-mult