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Proof step 33 from proof cong-equiv (previous step)

n0 = (n1 + m0 * m + n0) % m, n0 < m ⊦ m = 0, m | n1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 34:
n1 % m = 0, n0 < m ⊦ m | n1

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ m0 < n ∧ n ≠ 0 → ((m + m0) % n = m0 ↔ m % n = 0); (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%n ≠ 0 ⊦ (m + n0 * n) % n = m % n; (s)