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Proof step 7 from proof div-eq-same-01 (previous step)

i / j = i ⊦ j = 0, i = 0, j = 1, j < 0 ∧ i < 0

User applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
⊦ 0 < j ∧ 0 ≤ i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j ∧ 0 ≤ i0)
Proof step 9:
i / j = i ⊦ 0 < j ∧ 0 ≤ i, j = 0, i = 0, j = 1, j < 0 ∧ i < 0
Proof step 20:
i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j ∧ 0 ≤ i0 ∧ 0 < j ∧ 0 ≤ i, i0 = i ⊦ j = 0, i = 0, j = 1, j < 0 ∧ i < 0

Rule argument

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

Additional information

Used terms: i, j, i0, j0