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Proof step 13 from proof gcd-abs-rew-02 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ i | abs(j) → i | j
Proof step 15:
¬ gcd(abs(i), abs(j)) | abs(j), gcd(abs(i), abs(j)) | abs(i), gcd(i, j) < gcd(abs(i), abs(j)) ⊦ gcd(i, j) = gcd(abs(i), abs(j)), gcd(abs(i), abs(j)) | j ∧ (i ≠ 0 ∨ j ≠ 0)

Rule argument

spec: int-gcd inst: name: div-abs-01 seq: ⊦ i | abs(j) → i | j subst: [j, i] → [j, gcd(abs(i), abs(j))] flags: rewrite, current

Additional information

Used simplifier rules: gcd-zero: ⊦ gcd(0, 0) = 0; Used terms: j, gcd(abs(i), abs(j))