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Proof step 8 from proof divides-add (previous step)

i | i * i0 + j0 ⊦ ∃ i0. i * i0 = j0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ i | j ↔ (∃ i0. i * i0 = j)
Proof step 10:
∃ i1. i * i1 = i * i0 + j0 ⊦ ∃ i0. i * i0 = j0

Rule argument

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

Additional information

Used terms: i * i0 + j0, i