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Proof step 9 from proof div-add1-negpos-neg-01 (previous step)

1 < j, j * (i2 + i1) + j0 < 0, i1 < 0, j0 < j, 0 < j0, 0 < j, 0 ≤ i2 * j + j0, 0 ≤ i2 ⊦ (j * (i2 + i1) + j0) / j = i1 + i2 + 1, j0 = 0

User applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
⊦ 0 < j ∧ i < 0 → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0)
Proof step 11:
j * (i2 + i1) + j0 = i0 * j + j1, 0 < j, 0 < j0, j0 < j, i1 < 0, j * (i2 + i1) + j0 < 0, 1 < j, abs(j1) < j, 0 ≤ i2, 0 ≤ i2 * j + j0, j1 ≤ 0, i0 ≤ 0 ⊦ i0 = i1 + i2 + 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] → [j * (i2 + i1) + j0, j, i, j1]

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j; (s) Used terms: j * (i2 + i1) + j0, j, i, j1