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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: prime-abs: ⊦ prime(j) → abs(j) = j;%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%abs-eqall: ⊦ abs(i * j) = abs(i) * abs(j); from specification int-mult%labs: ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); from specification intnat