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Proof step 7 from proof equality-02 (previous step)

i0 < 0, i * j0 = i0 * j, 0 < j, 0 < j0, i < 0 ⊦ rational(i, j) = rational(i0, j0)

User applied rule insert lemma

And obtained the following new goals

Rule argument

name: equality-pos subst: [i, j0, i0, j] → [~ i, j0, ~ i0, j]

Additional information

Used simplifier rules: ⊦ ~ i * j = ~(i * j);% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ ~ i = ~ j ↔ i = j;% ⊦ 0 < ~ i ↔ i < 0;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false); Used terms: ~ i, j0, ~ i0, j