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Proof step 10 from proof num-div-denom-01 (previous step)

0 < i ∧ i1 < 0, i1 < 0 ⊦ i = 0, i1 = 0, j = 0, (~(i / gcd(i, i1)) * j) / i1 / gcd(i, i1) + (i * j) / i1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
i1 < 0, 0 < i ⊦ i = 0, i1 = 0, j = 0, (i * j) / i1 = (i / gcd(i, i1) * j) / i1 / gcd(i, i1)

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ i + 0 = i;% ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + ~ j = i - j;%j ≠ 0 ⊦ ~ i / j = ~(i / j);% ⊦ i0 < 0 → i0 / gcd(i, i0) ≠ 0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ ~ i * j = ~(i * j);