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Proof step 21 from proof abs-pos (previous step)

0 < j ⊦ i = 0, j = 0, rational(abs(rational(i, j).num), rational(i, j).denom).num = rational(abs(i), j).num

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
0 < j ⊦ i = 0, j = 0, rational(abs(i) / gcd(i, j), j / gcd(i, j)).num = rational(abs(i), j).num

Rule argument

Additional information

Used simplifier rules: numerator-pos: ⊦ i ≠ 0 ∧ 0 < j → rational(i, j).num = i / gcd(i, j);%denominator-pos: ⊦ i ≠ 0 ∧ j ≠ 0 → rational(i, j).denom = abs(j) / gcd(i, j);% ⊦ 0 < j → abs(j) = j;% ⊦ abs(gcd(i, j)) = gcd(i, j);%j ≠ 0 ⊦ abs(i / j) = abs(i) / abs(j);% ⊦ gcd(i, j) = 0 ↔ i = 0 ∧ j = 0;