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Proof step 3 from proof div-minus-onediff (previous step)

0 < j, i < 0 ⊦ i % j = 0, i / j = (i + 1 - j) / j + 1

User applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
⊦ 0 < j ∧ i < 0 → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0)
Proof step 5:
i = i0 * j + j0, i < 0, 0 < j, abs(j0) < j, j0 ≤ 0, i0 ≤ 0 ⊦ i0 = (i + 1 - j) / j + 1, j0 = 0

Rule argument

spec: <toplevel> name: div-elim-negpos seq: ⊦ 0 < j ∧ i < 0 → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0) subst: [i, j, i0, j0] → [i, j, i0, j0] flags: <rotated> <all paths>

Additional information

Used terms: i, j, i0, j0