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Proof step 10 from proof div-div-pred-01 (previous step)

abs(j0) < 2, i0 ≤ 0, j0 ≤ 0, i0 * 2 + j0 < ~1, i0 * 2 + j0 < 0 ⊦ (i0 - 3) / 4 = (j0 + i0 * 2 - 6) / 8

User applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
⊦ 0 < j ∧ i < 0 → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0)
Proof step 12:
i0 - 3 = i1 * 4 + j1 ∧ j1 ≤ 0 ∧ abs(j1) < 4 ∧ i1 ≤ 0 ∧ 0 < 4 ∧ i0 - 3 < 0, abs(j0) < 2, i0 ≤ 0, j0 ≤ 0, i0 * 2 + j0 < ~1, i0 * 2 + j0 < 0 ⊦ i1 = (j0 + i0 * 2 - 6) / 8

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] → [i0 - 3, 4, i, j] flags: current, []

Additional information

Used simplifier rules: sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%numlsledel: j0 ≠ 0, j < j0 ⊦ i ≤ (* j) → i < (* j0); from specification int-abs Used terms: i0 - 3, 4, i, j