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Proof step 12 from proof gcd-zero-01-h (previous step)

gcd(m, n1) = 0, n1 ≤ m, ∀ n0, m0. n0 < n1 + m → gcd(m0, n0) = 0 ∧ m0 < n0 → n0 = 0 ⊦ n1 + m = 0, n1 = 0

User applied rule apply induction

And closed the goal!

Rule argument

subst: [n0, m0] → [m, n1]

Additional information

Used simplifier rules: gcd-a: ⊦ gcd(m, gcd(n, n0)) = gcd(gcd(m, n), n0);%gcd-sym-eq: ⊦ n0 = gcd(m, n) → (n0 ≠ gcd(n, m) ↔ false);%gcd-id: ⊦ gcd(m, m) = m;%gcd-zero: ⊦ gcd(0, n) = n;%gcd-a: ⊦ gcd(m, gcd(n, n0)) = gcd(gcd(m, n), n0);%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%anti: ⊦ m ≤ n → (n ≤ m ↔ m = n); from specification nat%le-add: ⊦ m + m0 ≤ m ↔ m0 = 0; from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: m, n1