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Proof step 23 from proof byte-law-righthift-not (previous step)

i2b(bits2int(sublist(0, n, int2bits(128)))) = i2b ((# int2bits(~128) ≤ i→n(bits2int(bitsand(int2bits(1), int2bits(31)))) ⊃ ~1 ;bits2int(sublist(0, # int2bits(~128) - i→n(bits2int(bitsand(int2bits(1), int2bits(31)))), int2bits(~128))) )), # int2bits(128) = n + i→n(bits2int(bitsand(int2bits(1), int2bits(31)))), i→n(bits2int(bitsand(int2bits(1), int2bits(31)))) < # int2bits(128) ⊦ n = 0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 1

Additional information