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Proof step 2 from proof div-abs-02 (previous step)

i | abs(j) ⊦ i | j

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-gcd inst: name: divides seq: ⊦ i | j ↔ (∃ i0. i * i0 = j) subst: [j, i] → [abs(j), i] flags: rewrite, current

Additional information

Used terms: abs(j), i