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Proof step 12 from proof gcd-le-both-h (previous step)

n1 ≤ m, ∀ 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(n1, m) ∧ ¬ n1 + m < gcd(n1, m)

User applied rule apply induction

And closed the goal!

Rule argument

subst: [n0, m0] → [m, n1]

Additional information

Used simplifier rules: gcd-id: ⊦ gcd(m, m) = m;% ⊦ m < n ↔ ¬ n ≤ m; (ss)%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%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 → (n ≤ m ↔ m = n); (s)% ⊦ m ≤ m; (s)% ⊦ m + m = 2 * m; (s)% ⊦ m0 * m < m ↔ m0 = 0 ∧ m ≠ 0; (ws)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult% ⊦ m + n ≤ m ↔ n = 0; (s) Used terms: m, n1