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Proof step 6 from proof gcd-div-rest (previous step)

gcd(i, j) | i, i = i1 * gcd(i, j) + j0, abs(j0) < gcd(i, j) ⊦ j0 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ i | j ↔ (∃ i0. i * i0 = j)
Proof step 8:
∃ i0. gcd(i, j) * i0 = i, i = i1 * gcd(i, j) + j0, abs(j0) < gcd(i, j) ⊦ j0 = 0

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)