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Proof step 28 from proof gcd-rec-sub-03 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 29:
⊦ gcd(i, j) = gcd(j, i)
Proof step 30:
0 < i0 - abs(i) ∧ i0 - abs(i) < abs(i0), abs(i) < i0, i < 0 ⊦ i0 < 0, gcd(abs(i), i0 - abs(i)) = gcd(abs(i), abs(i0)), 0 < i ∧ i < i0 + i

Rule argument

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

Additional information

Used terms: abs(i), abs(i0)