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Proof step 107 from proof shiftright-rec-8-neg (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 108:
(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) / 2 >> (j - 9), ((i - 1) / 2) % 256 = 0, j = 8, i % 256 = 0, ((i - 1) / 2) / 256 = (i / 256 - 2) / 2 + 1

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i ≤ ~1 ↔ i < 0;% ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ (* i) + i0 - (* j) = (i - j) + i0;% ⊦ i + (j - i0) = i + j - i0;%0 < j ⊦ i / (* j) < 0 ↔ i + j ≤ 0;% ⊦ i - j = i0 ↔ i = i0 + j;