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Proof step 4 from proof exdivmod (previous step)

0 ≤ i ∧ 0 < j → (∃ i0, j0. i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j) ⊦ j = 0, ∃ i0, j0. i = i0 * j + j0 ∧ abs(j0) < abs(j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information