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Proof step 15 from proof cong-equiv (previous step)

n0 = m1 * m + n2 ∧ n2 < m ∧ m ≠ 0, n1 + n0 = m0 * m + n2, n2 < m ⊦ n1 = 0, m | n1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
n1 + m1 * m = m0 * m, n2 < m ⊦ n1 = 0, m | n1

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; from specification nat-basic1%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2