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Proof step 10 from proof byte-law-righthift-not (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 1

Additional information