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Proof step 14 from proof mult-def-03 (previous step)

0 < j1 ∧ 0 < abs(j2), j2 < 0, 0 < j1, j2 < 0 ⊦ j2 = 0, rational(i1 * ~ i2, j1 * abs(j2)) = rational(i1 * i2, j1 * j2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
0 < j1, j2 < 0 ⊦ j2 = 0, rational(~(i2 * i1), ~(j2 * j1)) = rational(i1 * i2, j1 * j2)

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ 0 < ~ i ↔ i < 0;% ⊦ i < 0 → abs(i) = ~ i;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ ~ i * j = ~(i * j);