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Proof step 6 from proof square-neq-any-pos-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
j * j = i * j0 * j0, gcd(j, j0) = 1, 0 < j0, 0 < j, 0 < i, 0 ≤ j0, ∀ i0, j, j0. i→n(i0) < i→n(i) → j * j = i0 * j0 * j0 ∧ gcd(j, j0) = 1 ∧ 0 < j0 ∧ 0 < j ∧ 0 < i0 → (∃ i1. i0 = i1 * i1) ⊦ prime(i), ∃ i0. i = i0 * i0

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs% ⊦ i < j → ¬ j < i; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)