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Proof step 20 from proof gcd-rec-sub-03 (previous step)

abs(i) < i0, i < 0 ⊦ i0 < 0, gcd(abs(i), i0 - abs(i)) = gcd(i0 + i, i0), 0 < i ∧ i < i0 + i

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 21:
⊦ gcd(i, j) = gcd(abs(i), abs(j))
Proof step 22:
abs(i) < i0, i < 0 ⊦ i0 < 0, gcd(abs(i), i0 - abs(i)) = gcd(abs(i0 + i), abs(i0)), 0 < i ∧ i < i0 + i

Rule argument

spec: int-gcd inst: name: gcd-abs-rew-02 seq: ⊦ gcd(i, j) = gcd(abs(i), abs(j)) subst: [j, i] → [i0, i0 + i] flags: rewrite, current

Additional information

Used terms: i0, i0 + i