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Proof step 10 from proof div-div (previous step)

¬ j < 0, (i / abs(j)) / abs(j0) = i / (abs(j) * abs(j0)), 0 ≤ i ⊦ (i / j) / j0 = i / (j * j0), j0 = 0, j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
(i / j) / abs(j0) = i / (j * abs(j0)), 0 ≤ i, 0 < j ⊦ (i / j) / j0 = i / (j * j0), j0 = 0

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1%Abs-pos: ⊦ ¬ j < 0 → abs(j) = j; from specification int-abs