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Proof step 11 from proof gcd-num-one (previous step)

i ≠ 0 ∧ 0 < i0 ⊦ i / gcd(i, i0) = 0, gcd(i / gcd(i, i0), rational(i, i0).denom) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
0 < i0 ⊦ gcd(i / gcd(i, i0), rational(i, i0).denom) = 1, i = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ i ≠ 0 → i / gcd(i, i0) ≠ 0;