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Proof step 76 from proof square-neq-any-pos-02 (previous step)

j * j = i * j0 * j0, gcd(j, j0) = 1, 0 < j0, 0 < j, 0 < i, 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) ⊦ 1 < i, prime(i), ∃ i0. i = i0 * i0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: gcd-same-01: ⊦ 0 < j → gcd(j, j) = j;%prime-small: ⊦ j < 2 → ¬ prime(j);%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%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%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%one-01: ⊦ i = 1 → i * j = j; from specification int-mult%bet: ⊦ 0 < i → (i ≤ 1 ↔ i = 1); from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs