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Proof step 102 from proof shiftright-rec-8-neg (previous step)

j0 * 256 >> (j - 1) = j0 >> (j - 9), 8 < j, 0 < j, j0 < 0, ∀ i, j0. i→n(j0) < i→n(j) → 8 ≤ j0 ∧ i < 0 ∧ 0 < j0 ∧ 0 ≤ j0 → i >> j0 = (i - 255) / 256 >> (j0 - 8) ⊦ j = 0, (512 * j0 + 1) / 512 = 1 + j0, (512 * j0 + 1) % 256 = 0, j = 8, j0 >> (j - 9) = ((512 * j0 + 1) / 512 - 1) >> (j - 9)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 103:
⊦ i % j = 0 ∧ i + i0 < 0 ∧ i0 % j ≠ 0 ∧ i < 0 ∧ 0 ≤ i0 ∧ 0 < j → (i + i0) / j = i / j + i0 / j + 1

Rule argument

seq: ⊦ i % j = 0 ∧ i + i0 < 0 ∧ i0 % j ≠ 0 ∧ i < 0 ∧ 0 ≤ i0 ∧ 0 < j → (i + i0) / j = i / j + i0 / j + 1 subst: [j, i0, i] → [512, 1, 512 * j0] rotate: F spec: int-div inst: name: div-add1-negpos-neg-01

Additional information

Used simplifier rules: ⊦ i + j = i + i0 ↔ j = i0;% ⊦ i / 1 = i;%j % j0 = 0, j ≠ 0, j0 ≠ 0 ⊦ (i * (* j0)) / (* j) = i / j / j0;%i0 % j = 0, j ≠ 0 ⊦ (i * (* i0)) % (* j) = 0;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ i < 0 ∧ j0 < j ∧ 0 ≤ j → i * j + j0 < 0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ 0 < i → (i * j < 0 ↔ j < 0);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false); Used terms: 512, 1, 512 * j0