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Proof step 11 from proof gcd-le-both-h (previous step)

m < gcd(n1, m), n1 ≤ m, n1 < gcd(n1, m), ∀ 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(n1, m) = 0

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;%gcd-a: ⊦ gcd(m, gcd(n, n0)) = gcd(gcd(m, n), n0);%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%anti: ⊦ m ≤ n → (n ≤ m ↔ m = n); from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%two-02: ⊦ m + m = 2 * m; from specification nat-mult%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%zero-06: ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0); from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat%le-add: ⊦ m + m0 ≤ m ↔ m0 = 0; from specification nat Used terms: m, n1