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Proof step 15 from proof abs-pos (previous step)

abs(i) ≠ 0 ∧ j ≠ 0, 0 < j ⊦ j = 0, i = 0, abs(rational(i, j)).num = rational(abs(i), j).num ∧ j / gcd(i, j) = abs(j) / gcd(abs(i), j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
0 < j ⊦ i = 0, j = 0, abs(rational(i, j)).num = rational(abs(i), j).num ∧ j / gcd(i, j) = j / gcd(abs(i), j)

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < j → abs(j) = j;% ⊦ abs(i) = 0 ↔ i = 0;