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Proof step 37 from proof Extsimple (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 38:
j1 / gcd(i1, j1) = j2 / gcd(i2, j2), i1 / gcd(i1, j1) = i2 / gcd(i2, j2), 0 < i1, j1 < 0, j2 < 0, i2 < 0 ⊦ i1 = 0, i2 = 0, j1 = 0, j2 = 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;% ⊦ i < j → ¬ j < i;% ⊦ i + 0 = i;% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + ~ j = i - j;%j ≠ 0 ⊦ ~ i / j = ~(i / j);% ⊦ gcd(i, j) = 0 ↔ i = 0 ∧ j = 0;% ⊦ i < 0 → abs(i) = ~ i;% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ ¬ i < j ↔ j ≤ i;