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Proof step 6 from proof square-neq-pos-h (previous step)

(i * i0) * i * i0 = i * j0 * j0, gcd(i * i0, j0) = 1, prime(i), 0 ≤ j0, 0 < i * i0 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-mult inst: name: eq-nonzero seq: ⊦ i ≠ 0 → (i * j = i * j0 ↔ j = j0) subst: [j0, j, i] → [j0 * j0, i0 * i * i0, i] flags: rewrite

Additional information

Used simplifier rules: gcd-zero-04: ⊦ gcd(i, 0) = abs(i);%gcd-com: ⊦ gcd(i, j) = gcd(j, i);%prime-small: ⊦ j < 2 → ¬ prime(j);%zero-left: ⊦ 0 * i = 0; from specification int-mult%absle: ⊦ 0 ≤ i → abs(i) = i; from specification int-abs%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: j0 * j0, i0 * i * i0, i