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Proof step 27 from proof div-eq-zero-01 (previous step)

j < 0, 0 ≤ i ⊦ i / j = 0 ↔ i + j < 0

User applied rule case distinction

And obtained the following new goals

Proof step 28:
i / j = 0, j < 0, 0 ≤ i ⊦ i + j < 0
Proof step 33:
i + j < 0, j < 0, 0 ≤ i ⊦ i / j = 0

Rule argument

pos: right 1

Additional information