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Proof step 5 from proof gcd-div-gcd (previous step)

gcd(i, j) = gcd(abs(i), abs(j)) ⊦ gcd(i / gcd(abs(i), abs(j)), j / gcd(abs(i), abs(j))) = 1, i = 0

User applied rule apply lemma

And obtained the following new goals

Proof step 6:
0 < i, 0 < j ⊦ gcd(i / gcd(i, j), j / gcd(i, j)) = 1
Proof step 7:
gcd(i, j) = gcd(abs(i), abs(j)), gcd(abs(i) / gcd(abs(i), abs(j)), abs(j) / gcd(abs(i), abs(j))) = 1 ⊦ gcd(i / gcd(abs(i), abs(j)), j / gcd(abs(i), abs(j))) = 1, i = 0

Rule argument

spec: <toplevel> name: gcd-div-gcd-pos seq: 0 < i, 0 < j ⊦ gcd(i / gcd(i, j), j / gcd(i, j)) = 1 subst: [i, j] → [abs(i), abs(j)]

Additional information

Used simplifier rules: gcd-zero-03: ⊦ i ≠ 0 → gcd(i, 0) = abs(i);%gcd-zero-04: ⊦ gcd(i, 0) = abs(i);% ⊦ 0 < abs(i) ↔ i ≠ 0; (s)% ⊦ abs(abs(i)) = abs(i); (s)% ⊦ abs(i) = 0 ↔ i = 0; (s)%j ≠ 0 ⊦ 0 / j = 0; (s)%j ≠ 0 ⊦ abs(i / abs(j)) = abs(i / j); (s)%i ≠ 0 ⊦ i / i = 1; (s) Used terms: abs(i), abs(j)