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Proof step 12 from proof le-div-to-mult (previous step)

(m1 + m) * m0 ≤ m1 * m0 + n0, ∃ m1. n0 + m1 = m0 ∧ m1 ≠ 0 ⊦ m0 = 0, m1 + m = 0, m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
(m1 + m) * (n0 + m2) ≤ m1 * (n0 + m2) + n0 ⊦ n0 + m2 = 0, m1 + m = 0, m = 0, m2 = 0

Rule argument

Additional information