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Proof step 17 from proof gcd-div-gcd-pos (previous step)

i = (i0 * i3) * gcd(i, j), j = i2 * gcd(j, i), 0 ≤ i, 1 < i0, 0 < i0, 0 < j, 0 < i, i0 | i2, 0 < i2, 0 < gcd(j, i), 0 ≤ j, 0 ≤ i2, 0 < i3, 0 ≤ i3, 0 < gcd(i, j) ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 18:
⊦ i | j ↔ (∃ i0. i * i0 = j)
Proof step 19:
i = (i0 * i3) * gcd(i, j), j = i2 * gcd(j, i), 0 ≤ i, 1 < i0, 0 < i0, 0 < j, 0 < i, ∃ i1. i0 * i1 = i2, 0 < i2, 0 < gcd(j, i), 0 ≤ j, 0 ≤ i2, 0 < i3, 0 ≤ i3, 0 < gcd(i, j) ⊦

Rule argument

spec: int-gcd inst: name: divides seq: ⊦ i | j ↔ (∃ i0. i * i0 = j) subst: [j, i] → [i2, i0] flags: rewrite, current

Additional information

Used terms: i2, i0