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Proof step 4 from proof exgcd-div (previous step)

gcd(i, j) | i ⊦ gcd(i, j) * i / gcd(i, j) = i

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ i | j ↔ (∃ i0. i * i0 = j)
Proof step 6:
∃ i0. gcd(i, j) * i0 = i ⊦ gcd(i, j) * i / gcd(i, j) = i

Rule argument

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

Additional information

Used terms: i, gcd(i, j)