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Proof step 1 from proof div-zero-02 (conclusion)

abs(i) < abs(j) ⊦ i / j = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
0 < j, 0 ≤ i, i < j ⊦ i / j = 0
Proof step 3:
¬(0 < j ∧ 0 ≤ i ∧ i < j), abs(i) < abs(j) ⊦ i / j = 0

Rule argument

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

Additional information

Used terms: j, i