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Proof step 14 from proof cong-equiv (previous step)

n1 + n0 = m0 * m + n2, n2 = n0 % m, n2 < m ⊦ m = 0, n1 = 0, m | n1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
n1 + n0 = m0 * m + n0 % m, n0 % m < m ⊦ n1 = 0, m | n1

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)