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Proof step 18 from proof Extsimple (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
i1 / gcd(i1, j1) + rational(i2, j2).num = 0, j1 / gcd(i1, j1) + abs(j2) / gcd(i2, j2) = 0, i1 < 0, j1 < 0 ⊦ i1 = 0, i2 = 0, j1 = 0, j2 = 0, ¬ 0 < i2 ∧ ¬ 0 < j2, i2 / gcd(i2, j2) + i1 / gcd(i1, j1) = 0 ∧ ¬ i2 < 0 ∧ ¬ j2 < 0, ¬ i2 < 0 ∧ ¬ 0 < j2

Rule argument

Additional information

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