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Proof step 8 from proof gcd-le-both-h (previous step)

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

User applied rule apply induction

And obtained the following new goals

Proof step 9:
∀ n0, m0. n0 < n1 + m → m0 < n0 → m0 = 0 ∨ n0 = 0 ∨ ¬ m0 < gcd(m0, n0) ∧ ¬ n0 < gcd(m0, n0) ⊦ m = 0, n1 = 0, m < n1, ¬ m < gcd(m, n1) ∧ ¬ n1 + m < gcd(m, n1)

Rule argument

subst: [n0, m0] → [n1, m]

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ ¬ n < n; (s)% ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + n ≤ m ↔ n = 0; (s) Used terms: n1, m