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Proof step 18 from proof square-neq-any-pos-02 (previous step)

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

User applied rule apply induction

And obtained the following new goals

Proof step 19:
j * j = (j1 * j1 * i) * j0 * j0, gcd(j, j0) = 1, 1 < j1 * j1 * i, 0 < j1 * j1 * i, 0 < j, 0 < j0, prime(j1), j1 * j1 ≤ j1 * j1 * i, 0 ≤ j0, ∀ i1, j, j0. i→n(i1) < i→n(j1 * j1 * i) → j * j = i1 * j0 * j0 ∧ gcd(j, j0) = 1 ∧ 0 < j0 ∧ 0 < j ∧ 0 < i1 → (∃ i2. i1 = i2 * i2) ⊦ i→n(i) < i→n(j1 * j1 * i), ∃ i1. j1 * j1 * i = i1 * i1
Proof step 25:
j * j = (j1 * j1 * i) * j0 * j0, gcd(j, j0) = 1, 1 < j1 * j1 * i, 0 < j1 * j1 * i, 0 < j, 0 < j0, prime(j1), j1 * j1 ≤ j1 * j1 * i, 0 ≤ j0, ∀ i1, j, j0. i→n(i1) < i→n(j1 * j1 * i) → j * j = i1 * j0 * j0 ∧ gcd(j, j0) = 1 ∧ 0 < j0 ∧ 0 < j ∧ 0 < i1 → (∃ i2. i1 = i2 * i2) ⊦ ∃ i1. j1 * j1 * i = i1 * i1, j * j = i * (j0 * j1) * j0 * j1 ∧ gcd(j, j0 * j1) = 1 ∧ 0 < j0 * j1 ∧ 0 < j ∧ 0 < i
Proof step 60:
j * j = (j1 * j1 * i) * j0 * j0, gcd(j, j0) = 1, 1 < j1 * j1 * i, 0 < j1 * j1 * i, 0 < j, 0 < j0, prime(j1), j1 * j1 ≤ j1 * j1 * i, 0 ≤ j0, ∃ i2. i = i2 * i2, ∀ i1, j, j0. i→n(i1) < i→n(j1 * j1 * i) → j * j = i1 * j0 * j0 ∧ gcd(j, j0) = 1 ∧ 0 < j0 ∧ 0 < j ∧ 0 < i1 → (∃ i2. i1 = i2 * i2) ⊦ ∃ i1. j1 * j1 * i = i1 * i1

Rule argument

subst: [i1, j, j0] → [i, j, j0 * j1]

Additional information

Used terms: i, j, j0 * j1