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Proof step 2 from proof bib-pos-high (previous step)

0 < i % 256, 127 < i % 256, 0 ≤ i ⊦ i % 256 = b2i(i2b(i)) + 256

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
~128 ≤ i, i ≤ 127 ⊦ b2i(i2b((* i))) = i
Proof step 4:
¬(~128 ≤ i ∧ i ≤ 127), 0 < i % 256, 127 < i % 256, 0 ≤ i ⊦ i % 256 = b2i(i2b(i)) + 256
Proof step 62:
~128 ≤ i ∧ i ≤ 127, 0 < i % 256, 127 < i % 256, 0 ≤ i ⊦ i % 256 = i + 256

Rule argument

name: num-ibbi rotate: F subst: [i] → [i]

Additional information

Used terms: i