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Proof step 2 from proof divides-mult (previous step)

⊦ i * j | j * i0 ↔ i | i0, j = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ i | j ↔ (∃ i0. i * i0 = j)
Proof step 4:
⊦ (∃ i1. (i * j) * i1 = j * i0) ↔ i | i0, j = 0

Rule argument

spec: <toplevel> name: divides seq: ⊦ i | j ↔ (∃ i0. i * i0 = j) subst: [j, i] → [j * i0, i * j] flags: <all paths>

Additional information

Used terms: j * i0, i * j