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Proof step 25 from proof square-neq-any-pos-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: prime-not-neg-01: ⊦ j ≤ 0 → ¬ prime(j);%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs% ⊦ ¬ i * i < 0; (s)% ⊦ i ≤ 0 → (0 < i * j ↔ j < 0 ∧ i ≠ 0); (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ 0 < i → (0 < i * j ↔ 0 < j); (s)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult