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Proof step 3 from proof num-ibbi (previous step)

~128 ≤ i, i ≤ 127 ⊦ b2i(i2b(i)) = i

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ i ∈byte → b2i(i2b(i)) = i

Rule argument

name: bib-in rotate: F subst: [i] → [i]

Additional information

Used simplifier rules: inbyte: ⊦ i ∈byte ↔ ~128 ≤ i ∧ i ≤ 127;%j0 ≠ 0, j ≤ j0 ⊦ i ≤ (* j) → i ≤ (* j0); Used terms: i