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Proof step 1 from proof gcd-div-mult (conclusion)

⊦ i ≠ 0 ∧ j ≠ 0 → i / gcd(i, j) * gcd(i, j) = i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ i / gcd(i, j) * gcd(i, j) = i, j = 0, i = 0

Rule argument

Additional information