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Proof step 8 from proof gcd-num-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
0 < i, i0 < 0 ⊦ i = 0, gcd(i / gcd(i, i0), rational(i, i0).denom) = 1

Rule argument

Additional information

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