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Proof step 150 from proof shiftright-rec-8-neg (previous step)

j1 < 0, (i1 * 256 + j1 - 1) >> (j - 1) = ((j1 - 1) / 256 + i1 - 1) >> (j - 9), 0 - j1 < abs(j1 - 1), 256 ≤ abs(j1 - 1), i1 < 0, i1 * 256 + j1 < 0, i1 < j, j1 < j, j1 ≤ 0, 8 < j, 0 < j, i1 * 256 + j1 < j, 0 - j1 < 256, ∀ i, j0. i→n(j0) < i→n(j) → 8 ≤ j0 ∧ i < 0 ∧ 0 < j0 ∧ 0 ≤ j0 → i >> j0 = (i - 255) / 256 >> (j0 - 8) ⊦ (i1 * 256 + j1 - 1) >> (j - 1) = (i1 - 1) >> (j - 9), (j1 - 1) % 256 = 0, j = 8, ((i1 * 256 + j1) * 2 - 1) % 256 = 0, (j1 - 1) / 256 = 0, j = 0, j1 = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 151:
(i1 * 256 + j1 - 1) >> (j - 1) = ((j1 - 1) / 256 + i1 - 1) >> (j - 9), 256 ≤ abs(j1 - 1), i1 < 0, i1 * 256 + j1 < 0, i1 < j, j1 < j, 8 < j, 0 < j, i1 * 256 + j1 < j, ~256 < j1, j1 < 0, 0 < abs(j1 - 1) + j1, ∀ i, j0. i→n(j0) < i→n(j) → 8 ≤ j0 ∧ i < 0 ∧ 0 < j0 ∧ 0 ≤ j0 → i >> j0 = (i - 255) / 256 >> (j0 - 8) ⊦ (i1 * 256 + j1 - 1) >> (j - 1) = (i1 - 1) >> (j - 9), (j1 - 1) % 256 = 0, j = 8, ((i1 * 256 + j1) * 2 - 1) % 256 = 0, (j1 - 1) / 256 = 0, j = 0, j1 = 1

Rule argument

Additional information

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