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Proof step 5 from proof le-div-to-mult (previous step)

m = m1 * m0 + n0, n0 < m0 ⊦ n ≤ m1 ↔ n * m0 ≤ m, m0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n0 < m0 ⊦ n ≤ m1 ↔ n * m0 ≤ m1 * m0 + n0

Rule argument

Additional information

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