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Proof step 2 from proof mod-eq-02 (previous step)

m0 < n ⊦ (m + m0) % n = m0 ↔ m % n = 0, n = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ m0 < n ∧ n ≠ 0 → ((m + m0) % n = m0 ↔ m % n = 0)

Rule argument

spec: <toplevel> name: mod-eq-01 seq: ⊦ m0 < n ∧ n ≠ 0 → ((m + m0) % n = m0 ↔ m % n = 0) subst: [n, m0, m] → [n, m0, m] flags: <all paths>

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) Used terms: n, m0, m