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Proof step 12 from proof bib-on-bits (previous step)

¬ i < ~1, 8 < # int2bits(i), ~128 ≤ i, i ≤ 0 ⊦ i = bits2int(lastn(8, int2bits(i)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
8 < # int2bits(i), i ≤ 0, ~1 ≤ i ⊦ i = bits2int(lastn(8, int2bits(i)))

Rule argument

Additional information

Used simplifier rules: i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;% ⊦ ¬ i < j ↔ j ≤ i;