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Proof step 20 from proof shiftright-rec-8-neg (previous step)

abs(i) < abs(256), i % 256 = 0, ~128 ≤ i, i < 0, ∀ i, j0. i→n(j0) < i→n(8) → 8 ≤ j0 ∧ i < 0 ∧ 0 < j0 ∧ 0 ≤ j0 → i >> j0 = (i - 255) / 256 >> (j0 - 8) ⊦ ~1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
i % 256 = 0, ~128 ≤ i, i < 0, ∀ i, j0. i→n(j0) < i→n(8) → 8 ≤ j0 ∧ i < 0 ∧ 0 < j0 ∧ 0 ≤ j0 → i >> j0 = (i - 255) / 256 >> (j0 - 8) ⊦

Rule argument

Additional information

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