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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, 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) ⊦ prime(i), ∃ i0. i = i0 * i0

User applied rule cut formula

And obtained the following new goals

Proof step 7:
1 < i, 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) → 0 < i0 ∧ 0 < j ∧ 0 < j0 ∧ gcd(j, j0) = 1 ∧ j * j = i0 * j0 * j0 → (∃ i1. i0 = i1 * i1) ⊦ prime(i), ∃ i0. i = i0 * i0
Proof step 76:
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) → 0 < i0 ∧ 0 < j ∧ 0 < j0 ∧ gcd(j, j0) = 1 ∧ j * j = i0 * j0 * j0 → (∃ i1. i0 = i1 * i1) ⊦ 1 < i, prime(i), ∃ i0. i = i0 * i0

Rule argument

fma: 1 < i

Additional information

Used formulas: 1 < i