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Proof step 4 from proof mod-mult (previous step)

∀ m, n. (m + n1 * n) % n = m % n ∨ n = 0 ⊦ n = 0, (m + n * n1) % n = m % n

User applied rule insert lemma

And obtained the following new goals

Proof step 5:
n ≠ 0 ⊦ (m + n) % n = m % n
Proof step 6:
((n1 * n + m) + n) % n = (n1 * n + m) % n, ∀ m, n. (m + n1 * n) % n = m % n ∨ n = 0 ⊦ n = 0, (m + n * n1) % n = m % n

Rule argument

spec: inst: name: mod-once seq: n ≠ 0 ⊦ (m + n) % n = m % n subst: [n, m] → [n, n1 * n + m] flags: current, []

Additional information

Used terms: n, n1 * n + m