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Proof step 2 from proof nummult-le-01 (previous step)

n < 0, 0 < m ⊦ m * m0 ≤ n ↔ m0 ≤ (n + 1 - m) / m

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: le-div-to-mult: m0 ≠ 0 ⊦ n ≤ m / m0 ↔ n * m0 ≤ m;% ⊦ ¬ m < 0; (s)% ⊦ 0 < m ↔ m ≠ 0; (s)% ⊦ m < n → n ≠ 0; (s)