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Proof step 31 from proof gcd-rec (previous step)

gcd(abs(j), j0) * i2 = abs(j), abs(i) = i0 * abs(j) + j0, gcd(abs(j), j0) * i1 = j0, gcd(i0 * abs(j) + j0, abs(j)) < gcd(abs(j), j0), 0 ≤ i0, 0 ≤ j0, j0 < abs(j) ⊦ i = 0, j = 0, gcd(i0 * abs(j) + j0, abs(j)) = gcd(abs(j), j0), gcd(abs(j), j0) | i0 * abs(j) + j0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 32:
⊦ i | j ↔ (∃ i0. i * i0 = j)
Proof step 33:
gcd(abs(j), j0) * i2 = abs(j), abs(i) = i0 * abs(j) + j0, gcd(abs(j), j0) * i1 = j0, gcd(i0 * abs(j) + j0, abs(j)) < gcd(abs(j), j0), 0 ≤ i0, 0 ≤ j0, j0 < abs(j) ⊦ i = 0, j = 0, gcd(i0 * abs(j) + j0, abs(j)) = gcd(abs(j), j0), ∃ i1. gcd(abs(j), j0) * i1 = i0 * abs(j) + j0

Rule argument

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

Additional information

Used terms: i0 * abs(j) + j0, gcd(abs(j), j0)