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Proof step 28 from proof div-eq-zero-01 (previous step)

0 ≤ i ⊦ i / j = 0 ↔ i < abs(j), 0 = j, 0 < i ∧ 0 < j

User applied rule case distinction

And obtained the following new goals

Proof step 29:
0 ≤ i ⊦ i / j = 0 ↔ i < abs(j), 0 = j, 0 < j

Rule argument

pos: right 3

Additional information

Used simplifier rules: zero: j ≠ 0 ⊦ 0 / j = 0;% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i ≤ j → (j ≤ i ↔ i = j); (s)% ⊦ 0 < abs(i) ↔ i ≠ 0; (s)