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Proof step 10 from proof square-neq-any-pos-02 (previous step)

j * j = i * j0 * j0, gcd(j, j0) = 1, j1 | i, prime(j1), j1 * j1 ≤ i, 0 < j0, 0 < j, 0 ≤ j0, 1 < i, 0 < i, ∀ i0, j, j0. i→n(i0) < i→n(i) → 0 < i0 ∧ 0 < j ∧ 0 < j0 ∧ gcd(j, j0) = 1 ∧ j * j = i0 * j0 * j0 → (∃ i1. i0 = i1 * i1) ⊦ ∃ i0. i = i0 * i0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ i | j ↔ (∃ i0. i * i0 = j)
Proof step 12:
j * j = i * j0 * j0, gcd(j, j0) = 1, ∃ i0. j1 * i0 = i, prime(j1), j1 * j1 ≤ i, 0 < j0, 0 < j, 0 ≤ j0, 1 < i, 0 < i, ∀ i0, j, j0. i→n(i0) < i→n(i) → 0 < i0 ∧ 0 < j ∧ 0 < j0 ∧ gcd(j, j0) = 1 ∧ j * j = i0 * j0 * j0 → (∃ i1. i0 = i1 * i1) ⊦ ∃ i0. i = i0 * i0

Rule argument

spec: int-gcd inst: name: divides seq: ⊦ i | j ↔ (∃ i0. i * i0 = j) subst: [j, i] → [i, j1] flags: rewrite, current

Additional information

Used terms: i, j1