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Proof step 21 from proof cong-equiv (previous step)

¬ m1 ≤ m0, n1 + m1 * m = m0 * m, n2 < m ⊦ n1 = 0, m * (m0 - m1) = n1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
n1 + m1 * m = m0 * m, m0 < m1, n2 < m ⊦ m * (m0 - m1) = n1, n1 = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat