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Proof step 4 from proof le-abs-numerator (previous step)

0 ≤ i ∧ 0 < j → i / j * j ≤ i ⊦ j = 0, 0 ≤ abs(i) + i / j ∧ i / j ≤ abs(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 ≤ i ∧ 0 < j → i / j * j ≤ i ⊦ j = 0, 0 ≤ abs(i) + i / j ∧ i / j ≤ abs(i)

Rule argument

Additional information