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Proof step 11 from proof isSquare-mult-aux (previous step)

prime(j0), j0 | j ∨ j0 | i * i, prime(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) ⊦ i = 0, isSquare(j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ prime(i) → (i | j * j0 ↔ i | j ∨ i | j0)
Proof step 13:
prime(j0), prime(j0), j0 | j ∨ j0 | i ∨ j0 | i, prime(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) ⊦ i = 0, isSquare(j)

Rule argument

spec: inst: name: prime-divides seq: ⊦ prime(i) → (i | j * j0 ↔ i | j ∨ i | j0) subst: [j0, j, i] → [i, i, j0] flags: rewrite, current

Additional information

Used terms: i, i, j0