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Proof step 2 from proof square-neq-any-pos-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
gcd(j, j0) = 1, j * j = i * j0 * j0, 0 < i, 0 < j, 0 < j0, ∀ i0, j, j0. i→n(i0) < i→n(i) → 0 < i0 ∧ 0 < j ∧ 0 < j0 ∧ gcd(j, j0) = 1 ∧ j * j = i0 * j0 * j0 → (∃ i1. i0 = i1 * i1) ⊦ ∃ i0. i = i0 * i0

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs