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Proof step 1 from proof inbyte-bib (conclusion)

⊦ i ∈byte ↔ b2i(i2b(i)) = i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
b2i(i2b(i)) = i ⊦ i ∈byte

Rule argument

Additional information

Used simplifier rules: inbyte: ⊦ i ∈byte ↔ ~128 ≤ i ∧ i ≤ 127;%bib-in-01: ⊦ ~128 ≤ i ∧ i ≤ 127 → b2i(i2b(i)) = i;