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Proof step 19 from proof isSquare-mult-aux (previous step)

j0 * j0 = j * i * i, prime(j0), 0 < j0, 0 < j, 0 < i, j0 | j, ∀ 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, j0 | i, isSquare(j)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: eq-pos: ⊦ 0 < i → (i * j = i * j0 ↔ j = j0); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%zero-gr: ⊦ 0 < i → (0 < i * j ↔ 0 < j); from specification int-mult%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%isSquare-square: ⊦ isSquare(i * i); from specification int-square Used terms: j, j0