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Proof step 16 from proof gcd-mult (previous step)

∀ 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

User applied rule apply induction

And closed the goal!

Rule argument

subst: [n1, m1, n2] → [n0, m, n4]

Additional information

Used simplifier rules: gcd-zero: ⊦ gcd(0, n) = n;%gcd-a: ⊦ gcd(m, gcd(n, n0)) = gcd(gcd(m, n), n0);% ⊦ m0 * (m + n) = m0 * m + m0 * n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ m + m = 2 * m; (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m0 * m < m0 * n ↔ m0 ≠ 0 ∧ m < n; (ws)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ 2 * n ≤ n ↔ n = 0; (s)% ⊦ m * 0 = 0; (s)% ⊦ m0 ≠ 0 → (m * m0 = n * m0 ↔ m = n); (s) Used terms: n0, m, n4