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Proof step 50 from proof shiftright-rec-8-neg (previous step)

(j0 * 256) / 2 >> (j - 1) = ((j0 * 256) / 2 - 255) / 256 >> (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 = 8, (j0 * 256) / 2 >> (j - 1) = (j0 - 1) / 2 >> (j - 9), j = 0, ((j0 * 256) / 2 - 255) / 256 = (j0 - 1) / 2

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 51:
⊦ i % 256 ≠ 0 ∧ i < 0 → (i - 255) / 256 = i / 256 - 1
Proof step 52:
¬(((j0 * 256) / 2) % 256 ≠ 0 ∧ (j0 * 256) / 2 < 0), (j0 * 256) / 2 >> (j - 1) = ((j0 * 256) / 2 - 255) / 256 >> (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 = 8, (j0 * 256) / 2 >> (j - 1) = (j0 - 1) / 2 >> (j - 9), j = 0, ((j0 * 256) / 2 - 255) / 256 = (j0 - 1) / 2
Proof step 63:
((j0 * 256) / 2) % 256 ≠ 0 ∧ (j0 * 256) / 2 < 0, (j0 * 256) / 2 >> (j - 1) = (((j0 * 256) / 2) / 256 - 1) >> (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 = 8, (j0 * 256) / 2 >> (j - 1) = (j0 - 1) / 2 >> (j - 9), j = 0, ((j0 * 256) / 2) / 256 - 1 = (j0 - 1) / 2

Rule argument

seq: ⊦ i % 256 ≠ 0 ∧ i < 0 → (i - 255) / 256 = i / 256 - 1 subst: [i] → [(j0 * 256) / 2] rotate: F spec: int-div inst: name: div-minus255-256

Additional information

Used terms: (j0 * 256) / 2