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Proof step 10 from proof one (previous step)

0 < ~ i, i < 0 ⊦ rational(~ i / gcd(~ i, ~ i), ~ i / gcd(~ i, ~ i)) = rational(1, 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
i < 0 ⊦ rational(~(i / gcd(i, i)), ~(i / gcd(i, i))) = rational(1, 1)

Rule argument

Additional information

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