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Proof step 10 from proof gcd-abs-rew-02 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-gcd inst: name: gcd-div1 seq: ⊦ gcd(i, j) | i subst: [j, i] → [abs(j), abs(i)] flags: rewrite, current

Additional information

Used terms: abs(j), abs(i)