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Proof step 26 from proof shiftright-rec-8-neg (previous step)

0 ≤ j, 0 < j, i < 0, 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) ⊦ j = 8, i >> j = (i - 255) / 256 >> (j - 8)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 27:
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 >> j = (i - 255) / 256 >> (j - 8), j = 8

Rule argument

Additional information

Used simplifier rules: i0 ≠ 0, i0 < i ⊦ (* i) < j → (* i0) < j;% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ i ≠ j → (i ≤ j ↔ i < j);