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Proof step 6 from proof gcd-le-both-h (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
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

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%nls-fst: ⊦ ¬ m < n → ¬ m + m0 < n; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l-01: ⊦ m < m + n ↔ n ≠ 0; from specification nat-basic2%lels-03: ⊦ m < n → m ≤ n; from specification nat