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Proof step 16 from proof square-neq-any-pos-02 (previous step)

∃ i1. j1 * i1 = i0, j * j = (j1 * i0) * j0 * j0, gcd(j, j0) = 1, prime(j1), j1 * j1 ≤ j1 * i0, 0 < j0, 0 < j, 0 ≤ j0, 1 < j1 * i0, 0 < j1 * i0, ∀ i1, j, j0. i→n(i1) < i→n(j1 * i0) → 0 < i1 ∧ 0 < j ∧ 0 < j0 ∧ gcd(j, j0) = 1 ∧ j * j = i1 * j0 * j0 → (∃ i2. i1 = i2 * i2) ⊦ ∃ i1. j1 * i0 = i1 * i1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
j * j = (j1 * j1 * i) * j0 * j0, gcd(j, j0) = 1, prime(j1), j1 * j1 ≤ j1 * j1 * i, 0 < j0, 0 < j, 0 ≤ j0, 1 < j1 * j1 * i, 0 < j1 * j1 * 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) ⊦ ∃ i1. j1 * j1 * i = i1 * i1

Rule argument

Additional information

Used simplifier rules: forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs