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Proof step 120 from proof shiftright-rec-8-neg (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 121:
(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) ⊦ i % 2 = 0, j = 0, (i / 2 - 1) >> (j - 1) = (i / 512 - 1) >> (j - 9), (i / 2 - 1) % 256 = 0, j = 8, i % 256 = 0, (i / 2 - 1) / 256 = i / 512

Rule argument

Additional information

Used simplifier rules: j ≠ 0, j0 ≠ 0 ⊦ (i / (* j)) / (* j0) = i / (j * j0);