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Proof step 27 from proof cong-equiv (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
⊦ m | n1 ↔ n % m = (n1 + n) % m, m = 0

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m ≤ m + n; (s)