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Proof step 4 from proof div-zero-02 (previous step)

i + j < 0, j < 0, 0 ≤ i ⊦ i / j = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
0 < j, i < 0, abs(i) < j ⊦ i / j = 0
Proof step 6:
¬(0 < j ∧ i < 0 ∧ abs(i) < j), i + j < 0, j < 0, 0 ≤ i ⊦ i / j = 0

Rule argument

spec: int-div inst: name: div-zero-01 seq: 0 < j, i < 0, abs(i) < j ⊦ i / j = 0 subst: [j, i] → [j, i] flags: rewrite, current

Additional information

Used terms: j, i