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Proof step 18 from proof gcd-one (previous step)

⊦ gcd(i, j) | j, gcd(i, j) = 1, ∃ i0. 1 < i0 ∧ i0 | i ∧ i0 | j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information