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Proof step 40 from proof square-neq-any-pos-02 (previous step)

j / gcd(j0 * j1, j) * j = (j0 * j1) / gcd(j0 * j1, j) * i * j0 * j1, (j / gcd(j0 * j1, j) * j) / gcd(j0 * j1, j) = ((j0 * j1) / gcd(j0 * j1, j) * i * j0 * j1) / gcd(j0 * j1, j), gcd(j, j0) = 1, 0 < j0, 0 < i, 0 < j1, 1 < j1 * j1 * i, 0 < j, prime(j1), 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) ⊦ gcd(j0 * j1, j) = 1, j = 0, j0 = 0, j1 = 0, ∃ i1. j1 * j1 * i = i1 * i1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 41:
⊦ i * j = j * i
Proof step 42:
j * j / gcd(j0 * j1, j) = (j0 * j1) / gcd(j0 * j1, j) * i * j0 * j1, (j * j / gcd(j0 * j1, j)) / gcd(j0 * j1, j) = ((j0 * j1) / gcd(j0 * j1, j) * i * j0 * j1) / gcd(j0 * j1, j), gcd(j, j0) = 1, 0 < j, 1 < j1 * j1 * i, 0 < j1, 0 < i, 0 < j0, prime(j1), 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) ⊦ gcd(j0 * j1, j) = 1, j = 0, j0 = 0, j1 = 0, ∃ i1. j1 * j1 * i = i1 * i1

Rule argument

spec: int-mult name: commutativity seq: ⊦ i * j = j * i subst: [j, i] → [j, j / gcd(j0 * j1, j)] flags: <all paths>

Additional information

Used terms: j, j / gcd(j0 * j1, j)