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Proof step 9 from proof gcd-zero-01-h (previous step)

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

User applied rule apply induction

And obtained the following new goals

Proof step 10:
gcd(m, n1) = 0, ∀ n0, m0. n0 < n1 + m → gcd(m0, n0) = 0 ∧ m0 < n0 → n0 = 0 ⊦ gcd(m, n1) = 0 ∧ m < n1, n1 + m = 0, n1 = 0

Rule argument

subst: [n0, m0] → [n1, m]

Additional information

Used simplifier rules: 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%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%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2 Used terms: n1, m