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Proof step 38 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) ⊦ ((i - 1) / 2 - 255) / 256 = ((i - 255) / 256 - 1) / 2, j = 0, (i - 1) / 2 >> (j - 1) = ((i - 255) / 256 - 1) / 2 >> (j - 9), j = 8

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 39:
⊦ i % 256 ≠ 0 ∧ i < 0 → (i - 255) / 256 = i / 256 - 1
Proof step 40:
¬(i % 256 ≠ 0 ∧ i < 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) ⊦ ((i - 1) / 2 - 255) / 256 = ((i - 255) / 256 - 1) / 2, j = 0, (i - 1) / 2 >> (j - 1) = ((i - 255) / 256 - 1) / 2 >> (j - 9), j = 8
Proof step 72:
i % 256 ≠ 0 ∧ i < 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) ⊦ ((i - 1) / 2 - 255) / 256 = (i / 256 - 1 - 1) / 2, j = 0, (i - 1) / 2 >> (j - 1) = (i / 256 - 1 - 1) / 2 >> (j - 9), j = 8

Rule argument

seq: ⊦ i % 256 ≠ 0 ∧ i < 0 → (i - 255) / 256 = i / 256 - 1 subst: [i] → [i] rotate: F spec: int-div inst: name: div-minus255-256

Additional information

Used terms: i