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Proof step 66 from proof square-neq-any-pos-02 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-mult name: eq-nonzero seq: ⊦ i ≠ 0 → (i * j = i * j0 ↔ j = j0) subst: [j0, j, i] → [j0 * j0 * i0, i * j1 * i, j1] flags: <all paths>

Additional information

Used simplifier rules: gcd-zero-04: ⊦ gcd(i, 0) = abs(i);%gcd-com: ⊦ gcd(i, j) = gcd(j, i);%prime-small: ⊦ j < 2 → ¬ prime(j);% ⊦ i * 0 = 0; (s)% ⊦ 0 < j → abs(j) = j; (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: j0 * j0 * i0, i * j1 * i, j1