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Proof step 8 from proof div-abs-01 (previous step)

j + i * i0 = 0, j < i * i0, j < 0, 0 < i * i0, j + 1 < i * i0 ⊦ i | j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ i | j ↔ (∃ i0. i * i0 = j)
Proof step 10:
j + i * i0 = 0, j < i * i0, j < 0, 0 < i * i0, j + 1 < i * i0 ⊦ ∃ i0. i * i0 = j

Rule argument

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

Additional information

Used terms: j, i