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Proof step 11 from proof gcd-rec-sub-03 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ 0 < i ∧ i < j → gcd(i, j) = gcd(j, j - i)
Proof step 13:
¬(0 < abs(i) ∧ abs(i) < abs(i0 + i)), i0 < 0, i < 0 ⊦ gcd(abs(i), abs(i0 + i)) = gcd(i0 + i, i0), 0 < i ∧ i < i0 + i
Proof step 15:
0 < abs(i) ∧ abs(i) < abs(i0 + i), i0 < 0, i < 0 ⊦ gcd(abs(i0 + i), abs(i0 + i) - abs(i)) = gcd(i0 + i, i0), 0 < i ∧ i < i0 + i

Rule argument

spec: int-gcd inst: name: gcd-rec-sub-02 seq: ⊦ 0 < i ∧ i < j → gcd(i, j) = gcd(j, j - i) subst: [j, i] → [abs(i0 + i), abs(i)] flags: rewrite, current

Additional information

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