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Proof step 8 from proof div-div-pred (previous step)

(i1 + 2) / 2 = i0 + 1, i1 < 0, i1 < ~3 ⊦ i0 / 2 = i1 / 4

User applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
⊦ 0 < j ∧ i < 0 → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0)
Proof step 10:
i1 + 2 = i2 * 2 + j0, i2 = i0 + 1, i1 < ~3, i1 < 0, abs(j0) < 2, 0 < 2, i1 + 2 < 0, j0 ≤ 0, i2 ≤ 0 ⊦ i0 / 2 = i1 / 4

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] → [i1 + 2, 2, i, j]

Additional information

Used simplifier rules: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i < j ↔ ¬ j ≤ i; (ss)%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs Used terms: i1 + 2, 2, i, j