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Proof step 6 from proof aligndown-is-n (previous step)

(n0 + n1) % n = 0, n < n0 + n1, n0 + n1 < 2 * n ⊦ n0 + n1 = n, n = 0, n = 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ n ≠ 0 → m % n = m - m / n * n
Proof step 8:
n0 + n1 - (n0 + n1) / n * n = 0, n0 + n1 < 2 * n, n < n0 + n1 ⊦ n0 + n1 = n, n = 1, n = 0

Rule argument

spec: nat-div name: mod-rew seq: ⊦ n ≠ 0 → m % n = m - m / n * n subst: [n, m] → [n, n0 + n1] flags: <all paths>

Additional information

Used terms: n, n0 + n1