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Proof step 3 from proof square-neq-any-pos-01 (previous step)

0 < i, 0 < j, 0 < j0, j * j = i * j0 * j0 ⊦ gcd(j, j0) = 1, ∃ i0. i = i0 * i0

User applied rule cut formula

And obtained the following new goals

Proof step 4:
(j * j) / gcd(j, j0) = (i * j0 * j0) / gcd(j, j0), 0 < i, 0 < j, 0 < j0, j * j = i * j0 * j0 ⊦ gcd(j, j0) = 1, ∃ i0. i = i0 * i0

Rule argument

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

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%neqf: ⊦ i / j ≠ i / i0 → j ≠ i0; (forward) from specification int-div Used formulas: (j * j) / gcd(j, j0) = (i * j0 * j0) / gcd(j, j0)