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Proof step 82 from proof Extsimple (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 83:
j1 / gcd(i1, j1) + abs(j2) / gcd(i2, j2) = 0, i1 / gcd(i1, j1) + abs(i2) / gcd(i2, j2) = 0, j1 < 0, i1 < 0, 0 < i2 ∧ 0 < j2 ∨ i2 < 0 ∧ j2 < 0 ⊦ j2 = 0, j1 = 0, i2 = 0, i1 = 0, i1 / gcd(i1, j1) + rational(i2, j2).num = 0

Rule argument

Additional information

Used simplifier rules: numerator-neg2: ⊦ i < 0 ∧ j < 0 → rational(i, j).num = abs(i) / gcd(i, j);% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ ~ i = j ↔ i + j = 0;%j ≠ 0 ⊦ ~ i / j = ~(i / j);% ⊦ gcd(i, j) = 0 ↔ i = 0 ∧ j = 0;% ⊦ i < 0 → abs(i) = ~ i;% ⊦ i < j → ¬ j < i;% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ ¬ i < j ↔ j ≤ i;