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Proof step 116 from proof shiftright-rec-8-neg (previous step)

2 ≠ 0, ∃ j0. i = j0 * 2, i / 2 >> (j - 1) = ((i / 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 / 2 >> (j - 1) = ((i / 256) / 2 - 1) >> (j - 9), (i / 2) % 256 = 0, j = 8, i % 256 = 0, (i / 2) / 256 = (i / 256) / 2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 117:
j0 >> (j - 1) = (j0 / 256 - 1) >> (j - 9), 0 < j, 8 < j, j0 < 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, j0 >> (j - 1) = (((j0 * 2) / 256) / 2 - 1) >> (j - 9), j0 % 256 = 0, j = 8, (j0 * 2) % 256 = 0, j0 / 256 = ((j0 * 2) / 256) / 2

Rule argument

Additional information

Used simplifier rules: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ 0 < i → (i * j < 0 ↔ j < 0);% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 1 * i = i;%j ≠ 0, j0 % j = 0 ⊦ (i * (* j0)) / (* j) = i * j0 / j;