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Proof step 8 from proof cong-equiv (previous step)

n = n1 + n0 ∧ n0 ≤ n, n0 < n, n0 < n ⊦ m = 0, m | n1 ↔ n % m = n0 % m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
⊦ m = 0, m | n1 ↔ (n1 + n0) % m = n0 % m, n1 = 0

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l-01: ⊦ m < m + n ↔ n ≠ 0; from specification nat-basic2%lels-03: ⊦ m < n → m ≤ n; from specification nat