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Proof step 15 from proof gcd-mult (previous step)

n0 * n = n3 + n0 * m, n = n4 + m, m < n, m ≤ n, ∀ n1, m1, n2. n1 * m1 + n1 * n2 < n0 * m + n0 * n → true → gcd(n1 * m1, n1 * n2) = n1 * gcd(m1, n2) ⊦ gcd(n0 * m, n3) = n0 * gcd(m, n4), n0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
∀ n1, m1, n2. n1 * m1 + n1 * n2 < n0 * m + n0 * (n4 + m) → true → gcd(n1 * m1, n1 * n2) = n1 * gcd(m1, n2) ⊦ gcd(n0 * m, n0 * n4) = n0 * gcd(m, n4), n0 = 0, n4 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ m < n → m ≤ n; (s)% ⊦ m + n = m + n0 ↔ n = n0; (s)%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m0 * (m + n) = m0 * m + m0 * n; (s)