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Proof step 13 from proof div-div-pred (previous step)

abs(j1) < 2, j0 ≤ 0, abs(j0) < 2, 2 * (i1 * 2 + j1) + j0 < ~3, j1 ≤ 0, i1 ≤ 0, i1 * 2 + j1 < 0, 2 * (i1 * 2 + j1) + j0 < 0 ⊦ i1 = (2 * (i1 * 2 + j1) + j0) / 4

User applied rule insert elim lemma

And obtained the following new goals

Proof step 14:
⊦ 0 < j ∧ i < 0 → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0)
Proof step 15:
2 * (i1 * 2 + j1) + j0 = i0 * 4 + j2 ∧ j2 ≤ 0 ∧ abs(j2) < 4 ∧ i0 ≤ 0 ∧ 0 < 4, abs(j1) < 2, j0 ≤ 0, abs(j0) < 2, 2 * (i1 * 2 + j1) + j0 < ~3, j1 ≤ 0, i1 ≤ 0, i1 * 2 + j1 < 0, 2 * (i1 * 2 + j1) + j0 < 0 ⊦ i1 = i0

Rule argument

spec: int-div inst: 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] → [2 * (i1 * 2 + j1) + j0, 4, i, j] flags: current, []

Additional information

Used simplifier rules: zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs Used terms: 2 * (i1 * 2 + j1) + j0, 4, i, j