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Proof step 84 from proof shiftright-rec-8-neg (previous step)

2 ≠ 0, j0 < 0, j0 * 256 >> (j - 1) = (j0 * 256 - 255) / 256 >> (j - 9), (i - 1) / 2 = j0 * 256, 8 < j, i < 0, 0 < j, ∀ i, j0. i→n(j0) < i→n(j) → 8 ≤ j0 ∧ i < 0 ∧ 0 < j0 ∧ 0 ≤ j0 → i >> j0 = (i - 255) / 256 >> (j0 - 8) ⊦ j = 8, i % 256 = 0, (j0 * 256 - 255) / 256 = ~((2 + ~(i / 256)) / 2), j = 0, j0 * 256 >> (j - 1) = ~((2 + ~(i / 256)) / 2) >> (j - 9)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 85:
⊦ i % j = 0 ∧ 0 ≤ i ∧ 0 ≤ i0 ∧ 0 < j → (i + i0) / j = i / j + i0 / j
Proof step 86:
2 % 2 = 0 ∧ 0 ≤ 2 ∧ 0 ≤ ~(i / 256) ∧ 0 < 2, 2 ≠ 0, j0 < 0, j0 * 256 >> (j - 1) = (j0 * 256 - 255) / 256 >> (j - 9), (i - 1) / 2 = j0 * 256, 8 < j, i < 0, 0 < j, ∀ i, j0. i→n(j0) < i→n(j) → 8 ≤ j0 ∧ i < 0 ∧ 0 < j0 ∧ 0 ≤ j0 → i >> j0 = (i - 255) / 256 >> (j0 - 8) ⊦ j = 8, i % 256 = 0, (j0 * 256 - 255) / 256 = ~(2 / 2 + ~(i / 256) / 2), j = 0, j0 * 256 >> (j - 1) = ~(2 / 2 + ~(i / 256) / 2) >> (j - 9)

Rule argument

seq: ⊦ i % j = 0 ∧ 0 ≤ i ∧ 0 ≤ i0 ∧ 0 < j → (i + i0) / j = i / j + i0 / j subst: [j, i0, i] → [2, ~(i / 256), 2] rotate: F spec: int-div inst: name: div-add1-pospos

Additional information

Used simplifier rules: ⊦ i + ~ j = i - j;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ ~ i = j ↔ i + j = 0;% ⊦ i ≤ ~ j ↔ i + j ≤ 0;% ⊦ i + 0 = i;% ⊦ 0 < j → (i / j ≤ 0 ↔ i < j);%j0 ≠ 0, j < j0 ⊦ i < (* j) → i < (* j0); Used terms: 2, ~(i / 256), 2