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Proof step 1 from proof Extsimple (conclusion)

⊦ i1 ≠ 0 ∧ i2 ≠ 0 ∧ j1 ≠ 0 ∧ j2 ≠ 0 → ( rational(i1, j1) = rational(i2, j2) ↔ abs(i1) / gcd(i1, j1) = abs(i2) / gcd(i2, j2) ∧ abs(j1) / gcd(i1, j1) = abs(j2) / gcd(i2, j2) ∧ ( (0 < i1 ∧ 0 < j1 ∨ i1 < 0 ∧ j1 < 0) ∧ (0 < i2 ∧ 0 < j2 ∨ i2 < 0 ∧ j2 < 0) ∨ (0 < i1 ∧ j1 < 0 ∨ i1 < 0 ∧ 0 < j1) ∧ (0 < i2 ∧ j2 < 0 ∨ i2 < 0 ∧ 0 < j2)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ ¬ i < j ↔ j ≤ i;