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Proof step 1 from proof divides-abs-01 (conclusion)

⊦ abs(i) | j ↔ i | j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ i | j ↔ (∃ i0. i * i0 = j)
Proof step 3:
⊦ (∃ i0. abs(i) * i0 = j) ↔ i | j

Rule argument

spec: int-gcd inst: name: divides seq: ⊦ i | j ↔ (∃ i0. i * i0 = j) subst: [j, i] → [j, abs(i)] flags: rewrite, current

Additional information

Used terms: j, abs(i)