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Proof step 3 from proof canon-any (previous step)

¬(0 ≤ i ∧ 0 < j) ⊦ 0 < j → rational(i, j) = rational(i / gcd(i, j), j / gcd(i, j))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
0 < j, i < 0 ⊦ rational(i, j) = rational(i / gcd(i, j), j / gcd(i, j))

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ ¬ i ≤ j ↔ j < i;