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Proof step 26 from proof gcd-abs-rew-02 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 27:
⊦ gcd(i, j) | j
Proof step 28:
gcd(abs(i), abs(j)) < gcd(i, j) ⊦ gcd(i, j) | i ∧ true ∧ (i ≠ 0 ∨ j ≠ 0)

Rule argument

spec: int-gcd inst: name: gcd-div2 seq: ⊦ gcd(i, j) | j subst: [j, i] → [j, i] flags: rewrite, current

Additional information

Used terms: j, i