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Proof step 59 from proof square-neq-any-pos-02 (previous step)

∃ i2. i = i2 * i2, 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) ⊦ ∃ 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;%gcd-mult-01: ⊦ gcd(i, i * j) = abs(i);%gcd-com: ⊦ gcd(i, j) = gcd(j, i);%one-01: ⊦ i = 1 → i * j = j; from specification int-mult%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%square-le: ⊦ i * i ≤ j * j ↔ abs(i) ≤ abs(j); from specification int-mult%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%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%square-eq: ⊦ i * i = j * j ↔ abs(i) = abs(j); from specification int-mult