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Proof step 2 from proof div-def-01 (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ 0 < i ∧ 0 < i0 ∧ 0 < j ∧ 0 < j0 → rational(i, j) / rational(i0, j0) = rational(i * j0, j * i0)
Proof step 4:
0 < abs(i) ∧ 0 < abs(i0) ∧ 0 < j ∧ 0 < j0 → rational(abs(i), j) / rational(abs(i0), j0) = rational(abs(i) * j0, j * abs(i0)), 0 < j, 0 < j0 ⊦ rational(i, j) / rational(i0, j0) = rational(i * j0, j * i0), i0 = 0

Rule argument

name: div-pos-def subst: [i, i0, j, j0] → [abs(i), abs(i0), j, j0]

Additional information

Used terms: abs(i), abs(i0), j, j0