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Proof step 1 from proof equality-02 (conclusion)

i * j0 = i0 * j, 0 < j, 0 < j0 ⊦ 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 terms: i, j0, i0, j