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Proof step 4 from proof square-neq-pos-h (previous step)

j * j = i * j0 * j0, gcd(j, j0) = 1, 0 < j, prime(i), i | j, 0 ≤ j0 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ i | j ↔ (∃ i0. i * i0 = j)
Proof step 6:
j * j = i * j0 * j0, gcd(j, j0) = 1, 0 < j, prime(i), 0 ≤ j0, ∃ i0. i * i0 = j ⊦

Rule argument

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

Additional information

Used terms: j, i