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Proof step 15 from proof num-div-denom (previous step)

gcd(i, i1) ≠ 0 ∧ i1 ≠ 0 ∧ i1 % gcd(i, i1) = 0, 0 < i1 ⊦ i = 0, i1 = 0, (i / gcd(i, i1) * gcd(i, i1)) / i1 = i / i1

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i ≠ 0 ∧ j ≠ 0 → i / gcd(i, j) * gcd(i, j) = i;% ⊦ gcd(i, j) = gcd(j, i);% ⊦ i ≠ 0 → i % gcd(i, j) = 0;