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Proof step 29 from proof shiftright-rec-8-neg (previous step)

i < 0 ∧ j ≠ 0, 0 < j, 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) ⊦ (i - 1) / 2 >> (abs(j) - 1) = (i - 255) / 256 >> (j - 8), j = 8

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 30:
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, j = 0, (i - 1) / 2 >> (j - 1) = (i - 255) / 256 >> (j - 8)

Rule argument

Additional information

Used simplifier rules: ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ 0 < j → abs(j) = j;