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Proof step 9 from proof le-div-to-mult (previous step)

n * m0 ≤ m1 * m0 + n0, n0 < m0, ∃ m0. m1 + m0 = n ∧ m0 ≠ 0 ⊦ m0 = 0, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
(m1 + m) * m0 ≤ m1 * m0 + n0, n0 < m0 ⊦ m0 = 0, m1 + m = 0, m = 0

Rule argument

Additional information