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

⊦ (i / j) / j0 = i / (j * j0), j0 = 0, j = 0

User applied rule insert lemma

And obtained the following new goals

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

Rule argument

spec: int-div inst: name: div-div-pos seq: ⊦ 0 ≤ i ∧ 0 < j ∧ 0 < j0 → (i / j) / j0 = i / (j * j0) subst: [i, j, j0] → [abs(i), abs(j), abs(j0)] flags: current, []

Additional information

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