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Proof step 111 from proof shiftright-rec-8-neg (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 112:
⊦ i % j ≠ 0 ∧ j0 = j - 1 ∧ i < 0 ∧ 0 < j → (i - j0) / j = i / j - 1
Proof step 113:
¬(i % 2 ≠ 0 ∧ 1 = 2 - 1 ∧ i < 0 ∧ 0 < 2), (i - 1) / 2 >> (j - 1) = (((i - 1) / 2) / 256 - 1) >> (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 = 0, (i - 1) / 2 >> (j - 1) = ((i / 256) / 2 - 1) >> (j - 9), ((i - 1) / 2) % 256 = 0, j = 8, i % 256 = 0, ((i - 1) / 2) / 256 = (i / 256) / 2
Proof step 120:
i % 2 ≠ 0 ∧ 1 = 2 - 1 ∧ i < 0 ∧ 0 < 2, (i / 2 - 1) >> (j - 1) = ((i / 2 - 1) / 256 - 1) >> (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 = 0, (i / 2 - 1) >> (j - 1) = ((i / 256) / 2 - 1) >> (j - 9), (i / 2 - 1) % 256 = 0, j = 8, i % 256 = 0, (i / 2 - 1) / 256 = (i / 256) / 2

Rule argument

seq: ⊦ i % j ≠ 0 ∧ j0 = j - 1 ∧ i < 0 ∧ 0 < j → (i - j0) / j = i / j - 1 subst: [j, j0, i] → [2, 1, i] rotate: F spec: int-div inst: name: div-minus-onediff

Additional information

Used terms: 2, 1, i