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Proof step 29 from proof square-neq-any-pos-02 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-gcd inst: name: gcd-div-shift seq: ⊦ i0 ≠ 0 → (i0 * j) / gcd(i0, j0) = i0 / gcd(i0, j0) * j subst: [j0, j, i0] → [j0 * j1, j, j] flags: rewrite, current

Additional information

Used simplifier rules: gcd-zero-04: ⊦ gcd(i, 0) = abs(i);%gcd-com: ⊦ gcd(i, j) = gcd(j, i);%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%abs-eqall: ⊦ abs(i * j) = abs(i) * abs(j); from specification int-mult%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%one-01: ⊦ i = 1 → i * j = j; from specification int-mult%div-zero-05: ⊦ i < j ∧ 0 < j ∧ 0 ≤ i → i / j = 0; from specification int-div Used terms: j0 * j1, j, j