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Proof step 27 from proof square-neq-any-pos-02 (previous step)

gcd(j, j0) = 1, j * j = i * (j0 * j1) * j0 * j1, 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 cut formula

And obtained the following new goals

Proof step 28:
(j * j) / gcd(j, j0 * j1) = (i * (j0 * j1) * j0 * j1) / gcd(j, j0 * j1), gcd(j, j0) = 1, j * j = i * (j0 * j1) * j0 * j1, 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

Rule argument

fma: (j * j) / gcd(j, j0 * j1) = (i * (j0 * j1) * j0 * j1) / gcd(j, j0 * j1)

Additional information

Used simplifier rules: neqf: ⊦ i / j ≠ i / i0 → j ≠ i0; (forward) from specification int-div Used formulas: (j * j) / gcd(j, j0 * j1) = (i * (j0 * j1) * j0 * j1) / gcd(j, j0 * j1)