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Proof step 26 from proof make (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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