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Proof step 3 from proof square-neq-any-pos-02 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ prime(i) ∧ gcd(j, j0) = 1 ∧ 0 < j ∧ 0 ≤ j0 → j * j ≠ i * j0 * j0
Proof step 5:
¬(prime(i) ∧ gcd(j, j0) = 1 ∧ 0 < j ∧ 0 ≤ j0), 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

spec: int-gcd inst: name: square-neq-pos-h seq: ⊦ prime(i) ∧ gcd(j, j0) = 1 ∧ 0 < j ∧ 0 ≤ j0 → j * j ≠ i * j0 * j0 subst: [j0, i, j] → [j0, i, j] flags: rewrite, current

Additional information

Used terms: j0, i, j