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Proof step 12 from proof num-div-denom (previous step)

i ≠ 0 ∧ 0 < i1 ⊦ i1 = 0, (i / gcd(i, i1)) / rational(i, i1).denom = i / i1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
0 < i1 ⊦ i = 0, i1 = 0, (i / gcd(i, i1)) / i1 / gcd(i, i1) = i / i1

Rule argument

Additional information

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