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Proof step 24 from proof isSquare-mult-aux (previous step)

max_prime_divisor(j0) | j0, j0 * j0 = j * i * i, 0 < j0, 0 < j, 0 < i, ∀ j1, j, i. i→n(j1) < i→n(j0) → j1 * j1 = j * i * i ∧ 0 < j1 ∧ 0 < j ∧ 0 < i → i = 0 ∨ isSquare(j) ⊦ prime(j0), i = 0, isSquare(j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 25:
⊦ i | j ↔ (∃ i0. i * i0 = j)
Proof step 26:
∃ i0. max_prime_divisor(j0) * i0 = j0, j0 * j0 = j * i * i, 0 < j0, 0 < j, 0 < i, ∀ j1, j, i. i→n(j1) < i→n(j0) → j1 * j1 = j * i * i ∧ 0 < j1 ∧ 0 < j ∧ 0 < i → i = 0 ∨ isSquare(j) ⊦ prime(j0), i = 0, isSquare(j)

Rule argument

spec: inst: name: divides seq: ⊦ i | j ↔ (∃ i0. i * i0 = j) subst: [j, i] → [j0, max_prime_divisor(j0)] flags: rewrite, current

Additional information

Used terms: j0, max_prime_divisor(j0)