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Proof step 5 from proof shiftright-byte-8-neg-01 (previous step)

0 ≤ j, 0 < j, b2i(by) < 0, 8 ≤ j ⊦ (b2i(by) - 255) / 256 >> (j - 8) = ~1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ i % 256 ≠ 0 ∧ i < 0 → (i - 255) / 256 = i / 256 - 1
Proof step 7:
b2i(by) % 256 ≠ 0 ∧ b2i(by) < 0, 0 ≤ j, 0 < j, b2i(by) < 0, 8 ≤ j ⊦ (b2i(by) / 256 - 1) >> (j - 8) = ~1

Rule argument

seq: ⊦ i % 256 ≠ 0 ∧ i < 0 → (i - 255) / 256 = i / 256 - 1 subst: [i] → [b2i(by)] rotate: F spec: int-div inst: name: div-minus255-256

Additional information

Used simplifier rules: byte-mod-100x: ⊦ b2i(by) % 256 = b2i(by);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ i < j → ¬ j < i;% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) Used terms: b2i(by)