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Proof step 23 from proof bib-pos-high (previous step)

# int2bits(j) = 8 + 1, 127 < j, 64 ≤ j, j < 256, 0 < j ⊦ j = 256 + b2i(i2b(bits2int(int2bits(j).rest))), 0 = j, int2bits(j).rest.first

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information