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Proof step 10 from proof gcd-zero-01-h (previous step)

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

User applied rule apply induction

And obtained the following new goals

Proof step 11:
gcd(m, n1) = 0, ∀ n0, m0. n0 < n1 + m → gcd(m0, n0) = 0 ∧ m0 < n0 → false ⊦ n1 = 0, gcd(m, n1) = 0 ∧ m < n1

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);% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m + m0 ≤ m ↔ m0 = 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: n1, m