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Proof step 24 from proof gcd-abs-rew-02 (previous step)

gcd(abs(i), abs(j)) < gcd(i, j) ⊦ gcd(i, j) | i ∧ gcd(i, j) | abs(j) ∧ (i ≠ 0 ∨ j ≠ 0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 25:
⊦ i | abs(j) ↔ i | j
Proof step 26:
gcd(abs(i), abs(j)) < gcd(i, j) ⊦ gcd(i, j) | i ∧ gcd(i, j) | j ∧ (i ≠ 0 ∨ j ≠ 0)

Rule argument

spec: int-gcd inst: name: div-abs-02 seq: ⊦ i | abs(j) ↔ i | j subst: [j, i] → [j, gcd(i, j)] flags: rewrite, current

Additional information

Used terms: j, gcd(i, j)