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Proof step 32 from proof shiftright-rec-8-neg (previous step)

(i - 1) / 2 >> (j - 1) = ((i - 1) / 2 - 255) / 256 >> (j - 9), 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, j = 0, (i - 1) / 2 >> (j - 1) = (i - 255) / 256 >> (j - 8)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 33:
⊦ i < 0 ∧ j ≠ 0 → i >> j = (i - 1) / 2 >> (abs(j) - 1)
Proof step 34:
(i - 255) / 256 < 0 ∧ j - 8 ≠ 0, (i - 1) / 2 >> (j - 1) = ((i - 1) / 2 - 255) / 256 >> (j - 9), 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, j = 0, (i - 1) / 2 >> (j - 1) = ((i - 255) / 256 - 1) / 2 >> (abs(j - 8) - 1)

Rule argument

name: shiftright-rec-neg rotate: F subst: [j, i] → [j - 8, (i - 255) / 256]

Additional information

Used simplifier rules: 0 < j ⊦ i / (* j) < 0 ↔ i + j ≤ 0;% ⊦ i + (j - i0) = i + j - i0;% ⊦ (* i) + i0 - (* j) = (i - j) + i0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j;% ⊦ i ≤ ~1 ↔ i < 0;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ i - j = i0 ↔ i = i0 + j; Used terms: j - 8, (i - 255) / 256