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Proof step 6 from proof gcd-div-gcd (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Proof step 6 from proof gcd-div-gcd (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information