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Proof step 15 from proof gcd-div-gcd-pos (previous step)

i = i1 * gcd(i, j), j = i2 * gcd(j, i), 0 < j, 0 < i, 0 < i0, 1 < i0, 0 < gcd(i, j), 0 < gcd(j, i), i0 | i1, i0 | i2, 0 ≤ i, 0 ≤ i1, 0 ≤ j, 0 ≤ i2 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ i | j ↔ (∃ i0. i * i0 = j)
Proof step 17:
i = i1 * gcd(i, j), j = i2 * gcd(j, i), 0 < gcd(j, i), 0 < gcd(i, j), 1 < i0, 0 < i0, 0 < i, 0 < j, i0 | i2, 0 ≤ i2, 0 ≤ j, 0 ≤ i1, 0 ≤ i, ∃ i2. i0 * i2 = i1 ⊦

Rule argument

spec: <toplevel> name: divides seq: ⊦ i | j ↔ (∃ i0. i * i0 = j) subst: [j, i] → [i1, i0] flags: <all paths>

Additional information

Used terms: i1, i0