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Proof step 3 from proof divides-abs-02 (previous step)

⊦ (∃ i0. i * i0 = abs(j) * j0) ↔ i | j * j0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ i | j ↔ (∃ i0. i * i0 = j)
Proof step 5:
⊦ (∃ i0. i * i0 = abs(j) * j0) ↔ (∃ i0. i * i0 = j * j0)

Rule argument

spec: int-gcd inst: name: divides seq: ⊦ i | j ↔ (∃ i0. i * i0 = j) subst: [j, i] → [j * j0, i] flags: rewrite, current

Additional information

Used terms: j * j0, i