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Proof step 5 from proof aligndown-is-n (previous step)

(n0 + n1) % n = 0, n0 + n1 < 2 * n, n ≤ n0 + n1 ⊦ n0 + n1 = n, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
(n0 + n1) % n = 0, n < n0 + n1, n0 + n1 < 2 * n ⊦ n0 + n1 = n, n = 0, n = 1

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s)%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ m0 * 1 = m0; (s)%m1 ≠ 0 ⊦ m + n < (* m1) * m ↔ n < (m1 - 1) * m; (s)% ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)