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Proof step 1 from proof divides-mult (conclusion)

⊦ j ≠ 0 → (i * j | j * i0 ↔ i | i0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ i | j ↔ (∃ i0. i * i0 = j)
Proof step 3:
⊦ j ≠ 0 → ((∃ i1. (i * j) * i1 = j * i0) ↔ i | i0)

Rule argument

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

Additional information

Used terms: j * i0, i * j