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Proof step 47 from proof bib-on-bits (previous step)

8 ≤ # int2bits(i) ⊦ ¬ bits2int(lastn(8, int2bits(i))) < ~128, b2i(i2b(bits2int(lastn(8, int2bits(i))))) = bits2int(lastn(8, int2bits(i))), ¬ i < ~128 ∧ ¬ 127 < i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 48:
bits2int(lastn(8, int2bits(i))) < ~128, 8 ≤ # int2bits(i), bits2int(lastn(8, int2bits(i))) < 0 ⊦ b2i(i2b(bits2int(lastn(8, int2bits(i))))) = bits2int(lastn(8, int2bits(i))), ¬ i < ~128 ∧ ¬ 127 < i

Rule argument

Additional information

Used simplifier rules: ⊦ i < (* j) ∧ j < 0 → i < 0; (forward)