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Proof step 9 from proof byte-law-righthift-not (previous step)

⊦ i = 128 ∧ j = 1 → 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 simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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