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Proof step 31 from proof Extsimple (previous step)

0 < i1 ∧ j1 < 0, ~(i1 / gcd(i1, j1)) = rational(i2, j2).num, j1 / gcd(i1, j1) + abs(j2) / gcd(i2, j2) = 0, j1 < 0 ⊦ j2 = 0, j1 = 0, i2 = 0, i1 = 0, abs(i1) / gcd(i1, j1) = abs(i2) / gcd(i2, j2) ∧ (¬ 0 < i1 ∧ (¬ i2 < 0 ∧ ¬ j2 < 0 ∨ ¬ 0 < i2 ∧ ¬ 0 < j2) ∨ ¬ i1 < 0 ∧ (¬ i2 < 0 ∧ ¬ 0 < j2 ∨ ¬ 0 < i2 ∧ ¬ j2 < 0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ i < j → ¬ j < i;% ⊦ 0 < j → abs(j) = j;% ⊦ ~ i = j ↔ i + j = 0;