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Proof step 49 from proof gcd-rec (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 50:
⊦ i | j ↔ (∃ i0. i * i0 = j)
Proof step 51:
abs(i) = i0 * abs(j) + j0, gcd(abs(i), abs(j)) * i1 = abs(j), j0 < abs(j), 0 ≤ j0, 0 ≤ i0, gcd(abs(j), j0) < gcd(abs(i), abs(j)), ∃ i0. gcd(abs(i), abs(j)) * i0 = abs(i) ⊦ j = 0, gcd(abs(i), abs(j)) | j0

Rule argument

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

Additional information

Used terms: abs(i), gcd(abs(i), abs(j))