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Proof step 14 from proof square-neq-any-pos-02 (previous step)

j1 | i0, j * j = (j1 * i0) * j0 * j0, gcd(j, j0) = 1, prime(j1), j1 * j1 ≤ j1 * i0, 0 < j0, 0 < j, 0 ≤ j0, 1 < j1 * i0, 0 < j1 * i0, ∀ i1, j, j0. i→n(i1) < i→n(j1 * i0) → 0 < i1 ∧ 0 < j ∧ 0 < j0 ∧ gcd(j, j0) = 1 ∧ j * j = i1 * j0 * j0 → (∃ i2. i1 = i2 * i2) ⊦ ∃ i1. j1 * i0 = i1 * i1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ i | j ↔ (∃ i0. i * i0 = j)
Proof step 16:
∃ i1. j1 * i1 = i0, j * j = (j1 * i0) * j0 * j0, gcd(j, j0) = 1, prime(j1), j1 * j1 ≤ j1 * i0, 0 < j0, 0 < j, 0 ≤ j0, 1 < j1 * i0, 0 < j1 * i0, ∀ i1, j, j0. i→n(i1) < i→n(j1 * i0) → 0 < i1 ∧ 0 < j ∧ 0 < j0 ∧ gcd(j, j0) = 1 ∧ j * j = i1 * j0 * j0 → (∃ i2. i1 = i2 * i2) ⊦ ∃ i1. j1 * i0 = i1 * i1

Rule argument

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

Additional information

Used terms: i0, j1