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Proof step 2 from proof div-elim2 (previous step)

i < j, i < 0, 0 < j, i + 1 < j ⊦ i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ 0 < j ∧ 0 < i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j)
Proof step 4:
0 < j ∧ 0 < ~ i → (~ i0 = ~ i / j ∧ ~ j0 = ~ i % j ↔ ~ i = ~ i0 * j + ~ j0 ∧ 0 ≤ ~ j0 ∧ ~ j0 < j), i < j, i < 0, 0 < j, i + 1 < j ⊦ i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j

Rule argument

spec: int-div inst: name: div-elim seq: ⊦ 0 < j ∧ 0 < i → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j) subst: [j, i, i0, j0] → [j, ~ i, ~ i0, ~ j0] flags: current, []

Additional information

Used terms: j, ~ i, ~ i0, ~ j0