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Proof step 41 from proof gcd-rec-sub-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 42:
⊦ gcd(i, j) = gcd(abs(i), abs(j))
Proof step 43:
0 < abs(i0) ∧ abs(i0) < abs(i), abs(i0) < abs(i), i < 0, 0 < i0 ⊦ gcd(abs(i), abs(i0 - abs(i))) = gcd(abs(i0), abs(i) - abs(i0)), i = 0

Rule argument

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

Additional information

Used terms: i0 - abs(i), i