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Proof step 25 from proof gcd-mult (previous step)

m = n4 + n ∧ n ≤ m, n0 * m = n3 + n0 * 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(n3, n0 * n) = n0 * gcd(n4, n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
∀ n1, m1, n2. n1 * m1 + n1 * n2 < n0 * (n4 + n) + n0 * n → true → gcd(n1 * m1, n1 * n2) = n1 * gcd(m1, n2) ⊦ gcd(n0 * n4, n0 * n) = n0 * gcd(n4, n), n0 = 0

Rule argument

Additional information

Used simplifier rules: lep: ⊦ m ≤ m + n; from specification nat%eq-01: ⊦ m + n = m + n0 ↔ n = n0; 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%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%dist: ⊦ m0 * (m + n) = m0 * m + m0 * n; from specification nat-mult