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Proof step 20 from proof div-eq-zero-01 (previous step)

0 < abs(j) + i, i < 0 ⊦ i / j = 0, 0 = j, i < j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
i < 0, j < 0, j < i, j ≤ i ⊦ i / j = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i + 0 = i; (s)% ⊦ i < j - i0 ↔ i + i0 < j; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + ~ j = i - j; (s)% ⊦ i < 0 → abs(i) = ~ i; (s)%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs% ⊦ i < j ↔ ¬ j ≤ i; (ss)