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Proof step 29 from proof bib-on-bits (previous step)

¬ 0 < i, # int2bits(i) < 8 ⊦ b2i(i2b(bits2int(lastn(8, int2bits(i))))) = bits2int(lastn(8, fillsgn(8, int2bits(i)))), i ∈byte

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward)% ⊦ ¬ i < j ↔ j ≤ i;