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Proof step 6 from proof shiftright-rec-8-neg (previous step)

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) ⊦ i >> 8 = (i - 255) / 256

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ i < ~(2 ^ (abs(j) - 1)) ∧ j ≠ 0 → i >> j = (i - (2 ^ abs(j) - 1)) / 2 ^ abs(j)
Proof step 8:
¬(i < ~(2 ^ (abs(8) - 1)) ∧ 8 ≠ 0), 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) ⊦ i >> 8 = (i - 255) / 256

Rule argument

name: shiftright-neg-div rotate: F subst: [j, i] → [8, i]

Additional information

Used terms: 8, i