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Proof step 12 from proof cong-equiv (previous step)

(n1 + n0) % m = n0 % m ⊦ m | n1, m = 0, n1 = 0

User applied rule insert elim lemma

And obtained the following new goals

Proof step 13:
n1 + n0 = m0 * m + n2 ∧ n2 < m, n2 = n0 % m ⊦ m | n1, m = 0, n1 = 0

Rule argument

spec: nat-div inst: name: div-elim seq: n ≠ 0 ⊦ m0 = m / n ∧ n0 = m % n ↔ m = m0 * n + n0 ∧ n0 < n subst: [m, n, m0, n0] → [n1 + n0, m, m0, n] flags:

Additional information

Used simplifier rules: n ≠ 0 ⊦ m0 = m / n ∧ n0 = m % n ↔ m = m0 * n + n0 ∧ n0 < n; (elim) Used terms: n1 + n0, m, m0, n