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Proof step 139 from proof shiftright-rec-8-neg (previous step)

¬ abs(j1 - 1) < abs(256), (j1 - 1) >> (j - 1) = ((j1 - 1) / 256 - 1) >> (j - 9), 8 < j, 0 < j, j1 < j, ~256 < j1, j1 < 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 = 0, (j1 - 1) >> (j - 1) = ~1 >> (j - 9), (j1 - 1) % 256 = 0, j = 8, (2 * j1 - 1) % 256 = 0, (j1 - 1) / 256 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 140:
(j1 - 1) >> (j - 1) = ((j1 - 1) / 256 - 1) >> (j - 9), 256 ≤ abs(j1 - 1), j1 < 0, ~256 < j1, j1 < j, 0 < j, 8 < j, ∀ i, j0. i→n(j0) < i→n(j) → 8 ≤ j0 ∧ i < 0 ∧ 0 < j0 ∧ 0 ≤ j0 → i >> j0 = (i - 255) / 256 >> (j0 - 8) ⊦ (j1 - 1) / 256 = 0, (2 * j1 - 1) % 256 = 0, j = 8, (j1 - 1) % 256 = 0, (j1 - 1) >> (j - 1) = ~1 >> (j - 9), j = 0, j1 = 1

Rule argument

Additional information

Used simplifier rules: ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ 0 < abs(i) ↔ i ≠ 0;% ⊦ 0 ≤ abs(i);% ⊦ i ≤ j ∧ 0 < i → 0 < j; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ ¬ i < j ↔ j ≤ i;