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Proof step 7 from proof gcd-le-both-h (previous step)

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

User applied rule apply induction

And obtained the following new goals

Proof step 8:
m < gcd(m, n1), ∀ n0, m0. n0 < n1 + m → m0 < n0 → n0 = 0 ∨ m0 = 0 ∨ ¬ m0 < gcd(m0, n0) ∧ ¬ n0 < gcd(m0, n0) ⊦ m < n1, n1 + m = 0, m = 0, n1 = 0, gcd(m, n1) = 0

Rule argument

subst: [n0, m0] → [n1, m]

Additional information

Used simplifier rules: 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 Used terms: n1, m