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Proof step 27 from proof shiftright-rec-8-neg (previous step)

0 < j, 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) ⊦ i >> j = (i - 255) / 256 >> (j - 8), j = 8

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 28:
⊦ i < 0 ∧ j ≠ 0 → i >> j = (i - 1) / 2 >> (abs(j) - 1)
Proof step 29:
i < 0 ∧ j ≠ 0, 0 < j, 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) ⊦ (i - 1) / 2 >> (abs(j) - 1) = (i - 255) / 256 >> (j - 8), j = 8

Rule argument

name: shiftright-rec-neg rotate: F subst: [j, i] → [j, i]

Additional information

Used simplifier rules: rightshift-zero: ⊦ i >> 0 = i;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false); Used terms: j, i