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Proof step 3 from proof div-div (previous step)

j | i, i | j, j ≤ 0, i ≤ 0 ⊦ i = j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ i | j ↔ (∃ i0. i * i0 = j)
Proof step 5:
j | i, i ≤ 0, j ≤ 0, ∃ i0. i * i0 = j ⊦ i = j

Rule argument

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

Additional information

Used terms: j, i