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Proof step 1 from proof bcompl-bib (conclusion)

⊦ i ∈byte → b2i(i2b(bcompl(i))) = bcompl(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
~128 ≤ i, i ≤ 127 ⊦ b2i(i2b(bcompl(i))) = bcompl(i)

Rule argument

Additional information

Used simplifier rules: inbyte: ⊦ i ∈byte ↔ ~128 ≤ i ∧ i ≤ 127;% ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward)