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Proof step 2 from proof aligndown-is-n (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
0 ≠ m ⊦ align↓(n, m) = (n % m = 0 ⊃ n;n / m * m)
Proof step 4:
n0 + n1 < 2 * n, n ≤ n0 + n1 ⊦ ((n0 + n1) % n = 0 ⊃ n0 + n1;(n0 + n1) / n * n) = n, n = 0

Rule argument

Additional information

Used simplifier rules: aligndown-def: 0 ≠ m ⊦ align↓(n, m) = (n % m = 0 ⊃ n;n / m * m);% ⊦ m0 ≠ 0 → (m0 < m0 * m ↔ 1 < m); (s)%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0); (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic