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Proof step 19 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, j1 * j1 * abs(i) ≤ abs(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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-mult inst: name: le-same-05 seq: ⊦ 1 < i → (i * abs(j) ≤ abs(j) ↔ j = 0) subst: [i, j] → [j1 * j1, i] flags: rewrite

Additional information

Used simplifier rules: zero-left: ⊦ 0 * i = 0; from specification int-mult%zero-neq-01: ⊦ 0 < i → (i * j = 0 ↔ j = 0); from specification int-mult%square-lez-01: ⊦ i * i ≤ 0 ↔ i = 0; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: j1 * j1, i