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Proof step 3 from proof div-eq-zero-01 (previous step)

⊦ i / j = 0 ↔ abs(i) < abs(j), 0 = j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ 0 < i ∧ 0 < j → (i / j = 0 ↔ i < j)
Proof step 5:
⊦ i / j = 0 ↔ abs(i) < abs(j), 0 = j, 0 < i ∧ 0 < j

Rule argument

spec: <toplevel> name: div-eq-zero seq: ⊦ 0 < i ∧ 0 < j → (i / j = 0 ↔ i < j) subst: [j, i] → [j, i] flags: <all paths>

Additional information

Used simplifier rules: ⊦ 0 < j → abs(j) = j; (s) Used terms: j, i